Program to find whether a given number is power of 2
Given a positive integer, write a function to find if it is a power of two or not.
Examples :
Input : n = 4 Output : Yes 22 = 4 Input : n = 7 Output : No Input : n = 32 Output : Yes 25 = 32
1. A simple method for this is to simply take the log of the number on base 2 and if you get an integer then the number is the power of 2
C++
// C++ Program to find whether a // no is power of two #include<bits/stdc++.h> using namespace std; // Function to check if x is power of 2 bool isPowerOfTwo( int n) { if (n==0) return false ; return ( ceil (log2(n)) == floor (log2(n))); } // Driver program int main() { isPowerOfTwo(31)? cout<< "Yes" <<endl: cout<< "No" <<endl; isPowerOfTwo(64)? cout<< "Yes" <<endl: cout<< "No" <<endl; return 0; } // This code is contributed by Surendra_Gangwar |
C
// C Program to find whether a // no is power of two #include<stdio.h> #include<stdbool.h> #include<math.h> /* Function to check if x is power of 2*/ bool isPowerOfTwo( int n) { if (n==0) return false ; return ( ceil (log2(n)) == floor (log2(n))); } // Driver program int main() { isPowerOfTwo(31)? printf ( "Yes\n" ): printf ( "No\n" ); isPowerOfTwo(64)? printf ( "Yes\n" ): printf ( "No\n" ); return 0; } // This code is contributed by bibhudhendra |
Java
// Java Program to find whether a // no is power of two class GFG { /* Function to check if x is power of 2*/ static boolean isPowerOfTwo( int n) { if (n== 0 ) return false ; return ( int )(Math.ceil((Math.log(n) / Math.log( 2 )))) == ( int )(Math.floor(((Math.log(n) / Math.log( 2 ))))); } // Driver Code public static void main(String[] args) { if (isPowerOfTwo( 31 )) System.out.println( "Yes" ); else System.out.println( "No" ); if (isPowerOfTwo( 64 )) System.out.println( "Yes" ); else System.out.println( "No" ); } } // This code is contributed by mits |
Python3
# Python3 Program to find # whether a no is # power of two import math # Function to check # Log base 2 def Log2(x): if x = = 0 : return false; return (math.log10(x) / math.log10( 2 )); # Function to check # if x is power of 2 def isPowerOfTwo(n): return (math.ceil(Log2(n)) = = math.floor(Log2(n))); # Driver Code if (isPowerOfTwo( 31 )): print ( "Yes" ); else : print ( "No" ); if (isPowerOfTwo( 64 )): print ( "Yes" ); else : print ( "No" ); # This code is contributed # by mits |
C#
// C# Program to find whether // a no is power of two using System; class GFG { /* Function to check if x is power of 2*/ static bool isPowerOfTwo( int n) { if (n==0) return false ; return ( int )(Math.Ceiling((Math.Log(n) / Math.Log(2)))) == ( int )(Math.Floor(((Math.Log(n) / Math.Log(2))))); } // Driver Code public static void Main() { if (isPowerOfTwo(31)) Console.WriteLine( "Yes" ); else Console.WriteLine( "No" ); if (isPowerOfTwo(64)) Console.WriteLine( "Yes" ); else Console.WriteLine( "No" ); } } // This code is contributed // by Akanksha Rai(Abby_akku) |
PHP
<?php // PHP Program to find // whether a no is // power of two // Function to check // Log base 2 function Log2( $x ) { return (log10( $x ) / log10(2)); } // Function to check // if x is power of 2 function isPowerOfTwo( $n ) { return ( ceil (Log2( $n )) == floor (Log2( $n ))); } // Driver Code if (isPowerOfTwo(31)) echo "Yes\n" ; else echo "No\n" ; if (isPowerOfTwo(64)) echo "Yes\n" ; else echo "No\n" ; // This code is contributed // by Sam007 ?> |
Javascript
<script> // javascript Program to find whether a // no is power of two /* Function to check if x is power of 2 */ function isPowerOfTwo(n) { if (n == 0) return false ; return parseInt( (Math.ceil((Math.log(n) / Math.log(2))))) == parseInt( (Math.floor(((Math.log(n) / Math.log(2)))))); } // Driver Code if (isPowerOfTwo(31)) document.write( "Yes<br/>" ); else document.write( "No<br/>" ); if (isPowerOfTwo(64)) document.write( "Yes<br/>" ); else document.write( "No<br/>" ); // This code is contributed by shikhasingrajput. </script> |
Output:
No Yes
Time Complexity: O(1)
Auxiliary Space: O(1)
2. Another solution is to keep dividing the number by two, i.e, do n = n/2 iteratively. In any iteration, if n%2 becomes non-zero and n is not 1 then n is not a power of 2. If n becomes 1 then it is a power of 2.
C++
#include <bits/stdc++.h> using namespace std; /* Function to check if x is power of 2*/ bool isPowerOfTwo( int n) { if (n == 0) return 0; while (n != 1) { if (n%2 != 0) return 0; n = n/2; } return 1; } /*Driver code*/ int main() { isPowerOfTwo(31)? cout<< "Yes\n" : cout<< "No\n" ; isPowerOfTwo(64)? cout<< "Yes\n" : cout<< "No\n" ; return 0; } // This code is contributed by rathbhupendra |
C
#include<stdio.h> #include<stdbool.h> /* Function to check if x is power of 2*/ bool isPowerOfTwo( int n) { if (n == 0) return 0; while (n != 1) { if (n%2 != 0) return 0; n = n/2; } return 1; } /*Driver program to test above function*/ int main() { isPowerOfTwo(31)? printf ( "Yes\n" ): printf ( "No\n" ); isPowerOfTwo(64)? printf ( "Yes\n" ): printf ( "No\n" ); return 0; } |
Java
// Java program to find whether // a no is power of two import java.io.*; class GFG { // Function to check if // x is power of 2 static boolean isPowerOfTwo( int n) { if (n == 0 ) return false ; while (n != 1 ) { if (n % 2 != 0 ) return false ; n = n / 2 ; } return true ; } // Driver program public static void main(String args[]) { if (isPowerOfTwo( 31 )) System.out.println( "Yes" ); else System.out.println( "No" ); if (isPowerOfTwo( 64 )) System.out.println( "Yes" ); else System.out.println( "No" ); } } // This code is contributed by Nikita tiwari. |
Python3
# Python program to check if given # number is power of 2 or not # Function to check if x is power of 2 def isPowerOfTwo(n): if (n = = 0 ): return False while (n ! = 1 ): if (n % 2 ! = 0 ): return False n = n / / 2 return True # Driver code if (isPowerOfTwo( 31 )): print ( 'Yes' ) else : print ( 'No' ) if (isPowerOfTwo( 64 )): print ( 'Yes' ) else : print ( 'No' ) # This code is contributed by Danish Raza |
C#
// C# program to find whether // a no is power of two using System; class GFG { // Function to check if // x is power of 2 static bool isPowerOfTwo( int n) { if (n == 0) return false ; while (n != 1) { if (n % 2 != 0) return false ; n = n / 2; } return true ; } // Driver program public static void Main() { Console.WriteLine(isPowerOfTwo(31) ? "Yes" : "No" ); Console.WriteLine(isPowerOfTwo(64) ? "Yes" : "No" ); } } // This code is contributed by Sam007 |
PHP
<?php // Function to check if // x is power of 2 function isPowerOfTwo( $n ) { if ( $n == 0) return 0; while ( $n != 1) { if ( $n % 2 != 0) return 0; $n = $n / 2; } return 1; } // Driver Code if (isPowerOfTwo(31)) echo "Yes\n" ; else echo "No\n" ; if (isPowerOfTwo(64)) echo "Yes\n" ; else echo "No\n" ; // This code is contributed // by Sam007 ?> |
Javascript
<script> /* Function to check if x is power of 2*/ function isPowerOfTwo(n) { if (n == 0) return 0; while (n != 1) { if (n%2 != 0) return 0; n = n/2; } return 1; } isPowerOfTwo(31)? document.write( "Yes" + "</br>" ): document.write( "No" + "</br>" ); isPowerOfTwo(64)? document.write( "Yes" ): document.write( "No" ); </script> |
Output :
No Yes
Time Complexity: O(log2n)
Auxiliary Space: O(1)
3. Another way is to use this simple recursive solution. It uses the same logic as the above iterative solution but uses recursion instead of iteration.
C++
// C++ program for above approach #include <bits/stdc++.h> using namespace std; // Function which checks whether a number is a power of 2 bool powerOf2( int n) { // base cases // '1' is the only odd number which is a power of 2(2^0) if (n == 1) return true ; // all other odd numbers are not powers of 2 else if (n % 2 != 0 || n == 0) return false ; // recursive function call return powerOf2(n / 2); } // Driver Code int main() { int n = 64; // True int m = 12; // False if (powerOf2(n) == 1) cout << "True" << endl; else cout << "False" << endl; if (powerOf2(m) == 1) cout << "True" << endl; else cout << "False" << endl; } // This code is contributed by Aditya Kumar (adityakumar129) |
C
// C program for above approach #include <stdbool.h> #include <stdio.h> // Function which checks whether a number is a power of 2 bool powerOf2( int n) { // base cases // '1' is the only odd number which is a power of 2(2^0) if (n == 1) return true ; // all other odd numbers are not powers of 2 else if (n % 2 != 0 || n == 0) return false ; // recursive function call return powerOf2(n / 2); } // Driver Code int main() { int n = 64; // True int m = 12; // False if (powerOf2(n) == 1) printf ( "True\n" ); else printf ( "False\n" ); if (powerOf2(m) == 1) printf ( "True\n" ); else printf ( "False\n" ); } // This code is contributed by Aditya Kumar (adityakumar129) |
Java
// Java program for // the above approach import java.util.*; class GFG{ // Function which checks // whether a number is a // power of 2 static boolean powerOf2( int n) { // base cases // '1' is the only odd number // which is a power of 2(2^0) if (n == 1 ) return true ; // all other odd numbers are // not powers of 2 else if (n % 2 != 0 || n == 0 ) return false ; // recursive function call return powerOf2(n / 2 ); } // Driver Code public static void main(String[] args) { //True int n = 64 ; //False int m = 12 ; if (powerOf2(n) == true ) System.out.print( "True" + "\n" ); else System.out.print( "False" + "\n" ); if (powerOf2(m) == true ) System.out.print( "True" + "\n" ); else System.out.print( "False" + "\n" ); } } // This code is contributed by Princi Singh |
Python3
# Python program for above approach # function which checks whether a # number is a power of 2 def powerof2(n): # base cases # '1' is the only odd number # which is a power of 2(2^0) if n = = 1 : return True # all other odd numbers are not powers of 2 elif n % 2 ! = 0 or n = = 0 : return False #recursive function call return powerof2(n / 2 ) # Driver Code if __name__ = = "__main__" : print (powerof2( 64 )) #True print (powerof2( 12 )) #False #code contributed by Moukthik a.k.a rowdyninja |
C#
// C# program for above approach using System; class GFG{ // Function which checks whether a // number is a power of 2 static bool powerOf2( int n) { // Base cases // '1' is the only odd number // which is a power of 2(2^0) if (n == 1) return true ; // All other odd numbers // are not powers of 2 else if (n % 2 != 0 || n == 0) return false ; // Recursive function call return powerOf2(n / 2); } // Driver code static void Main() { int n = 64; //True int m = 12; //False if (powerOf2(n)) { Console.Write( "True" + "\n" ); } else { Console.Write( "False" + "\n" ); } if (powerOf2(m)) { Console.Write( "True" ); } else { Console.Write( "False" ); } } } // This code is contributed by rutvik_56 |
Javascript
<script> // javascript program for // the above approach // Function which checks // whether a number is a // power of 2 function powerOf2(n) { // base cases // '1' is the only odd number // which is a power of 2(2^0) if (n == 1) return true ; // all other odd numbers are // not powers of 2 else if (n % 2 != 0 || n ==0) return false ; // recursive function call return powerOf2(n / 2); } // Driver Code //True var n = 64; //False var m = 12; if (powerOf2(n) == true ) document.write( "True" + "\n" ); else document.write( "False" + "\n" ); if (powerOf2(m) == true ) document.write( "True" + "\n" ); else document.write( "False" + "\n" ); // This code contributed by shikhasingrajput </script> |
True False
Time Complexity: O(log2n)
Auxiliary Space: O(log2n)
4. All power of two numbers has only a one-bit set. So count the no. of set bits and if you get 1 then the number is a power of 2. Please see Count set bits in an integer for counting set bits.
C++
#include <bits/stdc++.h> using namespace std; #define bool int /* Function to check if x is power of 2*/ bool isPowerOfTwo( int n) { /* First x in the below expression is for the case when * x is 0 */ int cnt = 0; while (n > 0) { if ((n & 1) == 1) { cnt++; } n = n >> 1; // keep dividing n by 2 using right // shift operator } if (cnt == 1) { // if cnt = 1 only then it is power of 2 return true ; } return false ; } /*Driver code*/ int main() { isPowerOfTwo(31) ? cout << "Yes\n" : cout << "No\n" ; isPowerOfTwo(64) ? cout << "Yes\n" : cout << "No\n" ; return 0; } // This code is contributed by devendra salunke |
Java
// Java program of the above approach import java.io.*; class GFG { // Function to check if x is power of 2 static boolean isPowerofTwo( int n) { int cnt = 0 ; while (n > 0 ) { if ((n & 1 ) == 1 ) { cnt++; // if n&1 == 1 keep incrementing cnt // variable } n = n >> 1 ; // keep dividing n by 2 using right // shift operator } if (cnt == 1 ) { // if cnt = 1 only then it is power of 2 return true ; } return false ; } public static void main(String[] args) { if (isPowerofTwo( 30 ) == true ) System.out.println( "Yes" ); else System.out.println( "No" ); if (isPowerofTwo( 128 ) == true ) System.out.println( "Yes" ); else System.out.println( "No" ); } } // This code is contributed by devendra salunke. |
C#
// C# program to check for power for 2 using System; class GFG { // Method to check if x is power of 2 static bool isPowerOfTwo( int n) { int cnt = 0; // initialize count to 0 while (n > 0) { // run loop till n > 0 if ((n & 1) == 1) { // if n&1 == 1 keep incrementing cnt // variable cnt++; } n = n >> 1; // keep dividing n by 2 using right // shift operator } if (cnt == 1) // if cnt = 1 only then it is power of 2 return true ; return false ; } // Driver method public static void Main() { Console.WriteLine(isPowerOfTwo(31) ? "Yes" : "No" ); Console.WriteLine(isPowerOfTwo(64) ? "Yes" : "No" ); } } // This code is contributed by devendra salunke |
Python3
# Python program to check if given # number is power of 2 or not # Function to check if x is power of 2 def isPowerOfTwo(n): cnt = 0 while n > 0 : if n & 1 = = 1 : cnt = cnt + 1 n = n >> 1 if cnt = = 1 : return 1 return 0 # Driver code if (isPowerOfTwo( 31 )): print ( 'Yes' ) else : print ( 'No' ) if (isPowerOfTwo( 64 )): print ( 'Yes' ) else : print ( 'No' ) # This code is contributed by devendra salunke |
Javascript
<script> // JavaScript code for the above approach // Function to check if x is power of 2 function isPowerofTwo(n) { let cnt = 0; while (n > 0) { if ((n & 1) == 1) { cnt++; // if n&1 == 1 keep incrementing cnt // variable } n = n >> 1; // keep dividing n by 2 using right // shift operator } if (cnt == 1) { // if cnt = 1 only then it is power of 2 return true ; } return false ; } // Driver code if (isPowerofTwo(30) == true ) document.write( "Yes" + "<br/>" ); else document.write( "No" + "<br/>" ); if (isPowerofTwo(128) == true ) document.write( "Yes" + "<br/>" ); else document.write( "No" + "<br/>" ); // This code is contributed by sanjoy_62. </script> |
Output :
No Yes
Time complexity : O(N)
Space Complexity : O(1)
5. If we subtract a power of 2 numbers by 1 then all unset bits after the only set bit become set; and the set bit becomes unset.
For example for 4 ( 100) and 16(10000), we get the following after subtracting 1
3 –> 011
15 –> 01111
So, if a number n is a power of 2 then bitwise & of n and n-1 will be zero. We can say n is a power of 2 or not based on the value of n&(n-1). The expression n&(n-1) will not work when n is 0. To handle this case also, our expression will become n& (!n&(n-1)) (thanks to https://www.geeksforgeeks.org/program-to-find-whether-a-no-is-power-of-two/Mohammad for adding this case).
Below is the implementation of this method.
Time complexity : O(1)
Space complexity : O(1)
C++
#include <bits/stdc++.h> using namespace std; #define bool int /* Function to check if x is power of 2*/ bool isPowerOfTwo ( int x) { /* First x in the below expression is for the case when x is 0 */ return x && (!(x&(x-1))); } /*Driver code*/ int main() { isPowerOfTwo(31)? cout<< "Yes\n" : cout<< "No\n" ; isPowerOfTwo(64)? cout<< "Yes\n" : cout<< "No\n" ; return 0; } // This code is contributed by rathbhupendra |
C
#include<stdio.h> #define bool int /* Function to check if x is power of 2*/ bool isPowerOfTwo ( int x) { /* First x in the below expression is for the case when x is 0 */ return x && (!(x&(x-1))); } /*Driver program to test above function*/ int main() { isPowerOfTwo(31)? printf ( "Yes\n" ): printf ( "No\n" ); isPowerOfTwo(64)? printf ( "Yes\n" ): printf ( "No\n" ); return 0; } |
Java
// Java program to efficiently // check for power for 2 class Test { /* Method to check if x is power of 2*/ static boolean isPowerOfTwo ( int x) { /* First x in the below expression is for the case when x is 0 */ return x!= 0 && ((x&(x- 1 )) == 0 ); } // Driver method public static void main(String[] args) { System.out.println(isPowerOfTwo( 31 ) ? "Yes" : "No" ); System.out.println(isPowerOfTwo( 64 ) ? "Yes" : "No" ); } } // This program is contributed by Gaurav Miglani |
Python3
# Python program to check if given # number is power of 2 or not # Function to check if x is power of 2 def isPowerOfTwo (x): # First x in the below expression # is for the case when x is 0 return (x and ( not (x & (x - 1 ))) ) # Driver code if (isPowerOfTwo( 31 )): print ( 'Yes' ) else : print ( 'No' ) if (isPowerOfTwo( 64 )): print ( 'Yes' ) else : print ( 'No' ) # This code is contributed by Danish Raza |
C#
// C# program to efficiently // check for power for 2 using System; class GFG { // Method to check if x is power of 2 static bool isPowerOfTwo ( int x) { // First x in the below expression // is for the case when x is 0 return x != 0 && ((x & (x - 1)) == 0); } // Driver method public static void Main() { Console.WriteLine(isPowerOfTwo(31) ? "Yes" : "No" ); Console.WriteLine(isPowerOfTwo(64) ? "Yes" : "No" ); } } // This code is contributed by Sam007 |
PHP
<?php // PHP program to efficiently // check for power for 2 // Function to check if // x is power of 2 function isPowerOfTwo ( $x ) { // First x in the below expression // is for the case when x is 0 return $x && (!( $x & ( $x - 1))); } // Driver Code if (isPowerOfTwo(31)) echo "Yes\n" ; else echo "No\n" ; if (isPowerOfTwo(64)) echo "Yes\n" ; else echo "No\n" ; // This code is contributed by Sam007 ?> |
Javascript
<script> // JavaScript program to efficiently // check for power for 2 /* Method to check if x is power of 2*/ function isPowerOfTwo (x) { /* First x in the below expression is for the case when x is 0 */ return x!=0 && ((x&(x-1)) == 0); } // Driver method document.write(isPowerOfTwo(31) ? "Yes" : "No" ); document.write( "<br>" +(isPowerOfTwo(64) ? "Yes" : "No" )); // This code is contributed by 29AjayKumar </script> |
Output :
No Yes
Time Complexity: O(1)
Auxiliary Space: O(1)
6. Another way is to use the logic to find the rightmost bit set of a given number.
C++
#include <iostream> using namespace std; /* Function to check if x is power of 2*/ bool isPowerofTwo( long long n) { if (n == 0) return 0; if ((n & (~(n - 1))) == n) return 1; return 0; } /*Driver code*/ int main() { isPowerofTwo(30) ? cout << "Yes\n" : cout << "No\n" ; isPowerofTwo(128) ? cout << "Yes\n" : cout << "No\n" ; return 0; } // This code is contributed by Sachin |
Java
// Java program of the above approach import java.io.*; class GFG { // Function to check if x is power of 2 static boolean isPowerofTwo( int n) { if (n == 0 ) return false ; if ((n & (~(n - 1 ))) == n) return true ; return false ; } public static void main(String[] args) { if (isPowerofTwo( 30 ) == true ) System.out.println( "Yes" ); else System.out.println( "No" ); if (isPowerofTwo( 128 ) == true ) System.out.println( "Yes" ); else System.out.println( "No" ); } } // This code is contributed by rajsanghavi9. |
Python3
# Python program of the above approach # Function to check if x is power of 2*/ def isPowerofTwo(n): if (n = = 0 ): return 0 if ((n & (~(n - 1 ))) = = n): return 1 return 0 # Driver code if (isPowerofTwo( 30 )): print ( 'Yes' ) else : print ( 'No' ) if (isPowerofTwo( 128 )): print ( 'Yes' ) else : print ( 'No' ) # This code is contributed by shivanisinghss2110 |
C#
// C# program of the above approach using System; public class GFG { // Function to check if x is power of 2 static bool isPowerofTwo( int n) { if (n == 0) return false ; if ((n & (~(n - 1))) == n) return true ; return false ; } public static void Main(String[] args) { if (isPowerofTwo(30) == true ) Console.WriteLine( "Yes" ); else Console.WriteLine( "No" ); if (isPowerofTwo(128) == true ) Console.WriteLine( "Yes" ); else Console.WriteLine( "No" ); } } // This code contributed by gauravrajput1 |
Javascript
<script> // javascript program of the above approach // Function to check if x is power of 2 function isPowerofTwo(n) { if (n == 0) return false ; if ((n & (~(n - 1))) == n) return true ; return false ; } if (isPowerofTwo(30) == true ) document.write( "Yes<br/>" ); else document.write( "No<br/>" ); if (isPowerofTwo(128) == true ) document.write( "Yes<br/>" ); else document.write( "No<br/>" ); // This code is contributed by umadevi9616 </script> |
No Yes
Time complexity : O(1)
Space complexity : O(1)
7. Brian Kernighan’s algorithm ( Efficient Method )
Approach :
As we know that the number which will be the power of two have only one set bit , therefore when we do bitwise and with the number which is just less than the number which can be represented as the power of (2) then the result will be 0 .
Example : 4 can be represented as (2^2 ) ,
(4 & 3)=0 or in binary (100 & 011=0)
Here is the code of the given approach :
C++
// C++ program to check whether the given number is power of // 2 #include <iostream> using namespace std; /* Function to check if x is power of 2*/ bool isPowerofTwo( long long n) { return (n != 0) && ((n & (n - 1)) == 0); } /*Driver code*/ int main() { isPowerofTwo(30) ? cout << "Yes\n" : cout << "No\n" ; isPowerofTwo(128) ? cout << "Yes\n" : cout << "No\n" ; return 0; } // This code is contributed by Suruchi Kumari |
Java
/*package whatever //do not write package name here */ import java.io.*; class GFG { /* Function to check if x is power of 2*/ public static boolean isPowerofTwo( long n) { return (n != 0 ) && ((n & (n - 1 )) == 0 ); } public static void main (String[] args) { if (isPowerofTwo( 30 )) { System.out.println( "Yes" ); } else { System.out.println( "No" ); } if (isPowerofTwo( 128 )) { System.out.println( "Yes" ); } else { System.out.println( "No" ); } } } // This code is contributed by akashish__ |
Python3
# Python3 program to check whether the given number is power of # 2 # Function to check if x is power of 2 def isPowerofTwo(n) : return (n ! = 0 ) and ((n & (n - 1 )) = = 0 ) # Driver code if __name__ = = "__main__" : if isPowerofTwo( 30 ) : print ( "Yes" ) else : print ( "No" ) if isPowerofTwo( 128 ) : print ( "Yes" ) else : print ( "No" ) # this code is contributed by aditya942003patil |
C#
using System; public class GFG{ /* Function to check if x is power of 2*/ static public bool isPowerofTwo( ulong n) { return (n != 0) && ((n & (n - 1)) == 0); } static public void Main (){ if (isPowerofTwo(30)) { System.Console.WriteLine( "Yes" ); } else { System.Console.WriteLine( "No" ); } if (isPowerofTwo(128)) { System.Console.WriteLine( "Yes" ); } else { System.Console.WriteLine( "No" ); } } } // This code is contributed by akashish__ |
Output :
No Yes
Time Complexity : O(1)
Auxiliary Space: O(1)
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