Given an integer **N**, the task is to check if there exists any permutation of the first **N** natural numbers **[1, N]** such that Bitwise AND of any pair of consecutive elements is not equal to **0**. If any such permutation exists, print **“Yes”**. Otherwise, print **“No”**.

**Examples:**

Input :5Output:YesExplanation:Permutation {2, 3, 1, 5, 4} satisfies the condition.

Input:4Output:NoExplanation:Since Bitwise AND of 4 and 3 is equal to 0, they cannot be placed adjacently. Similarly 2 and 4, 1 and 2, 1 and 4 cannot be placed adjacently. Therefore, no such permutation exists.

**Approach:** The problem can be solved based on the following observations:

- If
**N**is a power of two, then**N & (N – 1)**is equal to**0**. Therefore, no possible solution exists. - Otherwise, a permutation exists.

Therefore, to solve the problem, simply check if **N** is a power of **2** or not. If found to be false, print “**Yes**“. Otherwise, print “**No**“.

Below is the implementation of the above approach:

## C++

`// C++ Program to implement` `// the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to check if a permutation` `// of first N natural numbers exist` `// with Bitwise AND of adjacent` `// elements not equal to 0` `void` `check(` `int` `n)` `{` ` ` `// If n is a power of 2` ` ` `if` `((n & n - 1) != 0)` ` ` `cout << ` `"YES"` `<< endl;` ` ` `else` ` ` `cout << ` `"NO"` `<< endl;` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `n = 5;` ` ` `check(n);` ` ` `return` `0;` `}` |

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## Java

`// Java Program to implement` `// the above approach` `import` `java.util.*;` `class` `solution{` ` ` `// Function to check if a ` `// permutation of first N ` `// natural numbers exist` `// with Bitwise AND of adjacent` `// elements not equal to 0` `static` `void` `check(` `int` `n)` `{` ` ` `// If n is a power of 2` ` ` `if` `((n & n - ` `1` `) != ` `0` `)` ` ` `System.out.println(` `"YES"` `);` ` ` `else` ` ` `System.out.println(` `"NO"` `);` `}` `// Driver Code` `public` `static` `void` `main(String args[])` `{` ` ` `int` `n = ` `5` `;` ` ` `check(n);` `}` `}` `// This code is contributed by SURENDRA_GANGWAR` |

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## Python3

`# Python3 program to implement` `# the above approach` `# Function to check if a permutation` `# of first N natural numbers exist` `# with Bitwise AND of adjacent` `# elements not equal to 0` `def` `check(n):` ` ` ` ` `# If n is a power of 2` ` ` `if` `((n & n ` `-` `1` `) !` `=` `0` `):` ` ` `print` `(` `"YES"` `)` ` ` `else` `:` ` ` `print` `(` `"NO"` `)` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `n ` `=` `5` ` ` ` ` `check(n)` `# This code is contributed by bgangwar59` |

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## C#

`// C# Program to implement` `// the above approach` `using` `System;` `class` `solution{` ` ` `// Function to check if a ` `// permutation of first N ` `// natural numbers exist` `// with Bitwise AND of adjacent` `// elements not equal to 0` `static` `void` `check(` `int` `n)` `{` ` ` `// If n is a power of 2` ` ` `if` `((n & n - 1) != 0)` ` ` `Console.WriteLine(` `"YES"` `);` ` ` `else` ` ` `Console.WriteLine(` `"NO"` `);` `}` `// Driver Code` `public` `static` `void` `Main(String []args)` `{` ` ` `int` `n = 5;` ` ` `check(n);` `}` `}` `// This code is contributed by 29AjayKumar` |

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**Output**

YES

**Time Complexity: **O(1)**Auxiliary Space:** O(1)

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