Given an integer **N**. The task is to count the numbers of pairs of integers **A** and **B** such that **A + B + N = A ^ B ^ N** and A and B are less than N.

**Examples:**

Input:N = 2

Output:3

Explanation:-

For N = 2

2 XOR 0 XOR 0 = 2+0+0

2 XOR 0 XOR 1 = 2+0+1

2 XOR 0 XOR 2 != 2+0+2

2 XOR 1 XOR 0 = 2+1+0

2 XOR 1 XOR 1 != 2+1+1

2 XOR 1 XOR 2 != 2+1+2

2 XOR 2 XOR 0 != 2+2+0

2 XOR 2 XOR 1 != 2+2+1

2 XOR 2 XOR 2 != 2+2+2So (0, 0), (0, 1) and (1, 0) are the required pairs. So the output is 3.

Input:N = 4

Output:9

**Approach **:

To make the **sum** of three numbers equal to the **xor** of three number with one of the number given we can do following:-

- Represent the fixed number in binary form.
- Traverse the binary expansion of the fixed number.
- If you find a 1 there is only one condition i.e. you take the other two number’s binary bits as 0 and 0.
- If you find a 0 there will be three conditions i.e. either you can have binary bits as (0, 0), (1, 0)

or (0, 1).

- The following above triplets of bits will never go for a carry so they are valid.
- So the answer will be
**3^(number of zeros in binary representation)**.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the above approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Defining ull to unsigned long long int ` `typedef` `unsigned ` `long` `long` `int` `ull; ` ` ` `// Function to calculate power of 3 ` `ull calculate(` `int` `bit_cnt) ` `{ ` ` ` `ull res = 1; ` ` ` `while` `(bit_cnt--) { ` ` ` `res = res * 3; ` ` ` `} ` ` ` ` ` `return` `res; ` `} ` ` ` `// Function to return the count of the ` `// unset bit ( zeros ) ` `int` `unset_bit_count(ull n) ` `{ ` ` ` `int` `count = 0; ` ` ` `while` `(n) { ` ` ` ` ` `// Check the bit is 0 or not ` ` ` `if` `((n & 1) == 0) ` ` ` `count++; ` ` ` `// Right shifting ( dividing by 2 ) ` ` ` `n = n >> 1; ` ` ` `} ` ` ` `return` `count; ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` `ull n; ` ` ` `n = 2; ` ` ` ` ` `int` `count = unset_bit_count(n); ` ` ` ` ` `ull ans = calculate(count); ` ` ` ` ` `cout << ans << endl; ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// Java implementation of the approach ` `import` `java.util.*; ` ` ` `class` `GFG ` `{ ` ` ` `// Function to calculate power of 3 ` `static` `long` `calculate(` `int` `bit_cnt) ` `{ ` ` ` `long` `res = ` `1` `; ` ` ` `while` `(bit_cnt-- > ` `0` `) ` ` ` `{ ` ` ` `res = res * ` `3` `; ` ` ` `} ` ` ` ` ` `return` `res; ` `} ` ` ` `// Function to return the count of the ` `// unset bit ( zeros ) ` `static` `int` `unset_bit_count(` `long` `n) ` `{ ` ` ` `int` `count = ` `0` `; ` ` ` `while` `(n > ` `0` `) ` ` ` `{ ` ` ` ` ` `// Check the bit is 0 or not ` ` ` `if` `((n & ` `1` `) == ` `0` `) ` ` ` `count++; ` ` ` ` ` `// Right shifting ( dividing by 2 ) ` ` ` `n = n >> ` `1` `; ` ` ` `} ` ` ` `return` `count; ` `} ` ` ` `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ` `long` `n; ` ` ` `n = ` `2` `; ` ` ` ` ` `int` `count = unset_bit_count(n); ` ` ` ` ` `long` `ans = calculate(count); ` ` ` ` ` `System.out.println(ans); ` `} ` `} ` ` ` `// This code is contributed by PrinciRaj1992 ` |

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## Python3

`# Python3 implementation of the approach ` ` ` `# Function to calculate power of 3 ` `def` `calculate(bit_cnt): ` ` ` ` ` `res ` `=` `1` `; ` ` ` `while` `(bit_cnt > ` `0` `): ` ` ` `bit_cnt ` `-` `=` `1` `; ` ` ` `res ` `=` `res ` `*` `3` `; ` ` ` `return` `res; ` ` ` `# Function to return the count of the ` `# unset bit ( zeros ) ` `def` `unset_bit_count(n): ` ` ` ` ` `count ` `=` `0` `; ` ` ` `while` `(n > ` `0` `): ` ` ` ` ` `# Check the bit is 0 or not ` ` ` `if` `((n & ` `1` `) ` `=` `=` `0` `): ` ` ` `count ` `+` `=` `1` `; ` ` ` ` ` `# Right shifting ( dividing by 2 ) ` ` ` `n ` `=` `n >> ` `1` `; ` ` ` ` ` `return` `count; ` ` ` `# Driver Code ` `if` `__name__ ` `=` `=` `'__main__'` `: ` ` ` ` ` `n ` `=` `2` `; ` ` ` ` ` `count ` `=` `unset_bit_count(n); ` ` ` ` ` `ans ` `=` `calculate(count); ` ` ` ` ` `print` `(ans); ` ` ` `# This code contributed by Rajput-Ji ` |

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## C#

`// C# implementation of the approach ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` `// Function to calculate power of 3 ` `static` `long` `calculate(` `int` `bit_cnt) ` `{ ` ` ` `long` `res = 1; ` ` ` `while` `(bit_cnt-- > 0) ` ` ` `{ ` ` ` `res = res * 3; ` ` ` `} ` ` ` ` ` `return` `res; ` `} ` ` ` `// Function to return the count of the ` `// unset bit (zeros) ` `static` `int` `unset_bit_count(` `long` `n) ` `{ ` ` ` `int` `count = 0; ` ` ` `while` `(n > 0) ` ` ` `{ ` ` ` ` ` `// Check the bit is 0 or not ` ` ` `if` `((n & 1) == 0) ` ` ` `count++; ` ` ` ` ` `// Right shifting ( dividing by 2 ) ` ` ` `n = n >> 1; ` ` ` `} ` ` ` `return` `count; ` `} ` ` ` `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` ` ` `long` `n; ` ` ` `n = 2; ` ` ` ` ` `int` `count = unset_bit_count(n); ` ` ` ` ` `long` `ans = calculate(count); ` ` ` ` ` `Console.WriteLine(ans); ` `} ` `} ` ` ` `// This code is contributed by 29AjayKumar ` |

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**Output:**

3

**Time Complexity**: O(Number of unset_bits)

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