Given a positive integer **N**, the task is to find the largest integer **M** such that **0 <= M < N** and **XOR(M, N)** is an even number. If such a value of M cannot be obtained for given **N**, print **-1**.

**Examples:**

Input:N = 10

Output:8

Explanation:

(10 XOR 9) = 3, so M = 9 is not possible because 3 is not an even number.

(10 XOR 8) = 2, so M = 8 is possible as it is the largest possible value of M satisfying the necessary conditions.

Input:N = 5

Output:3

(5 XOR 4) = 1, so M = 4 is not possible because 1 is not an even number.

(5 XOR 3) = 6, so M = 3 is possible as 6 is an even number and 3 is the largest possible value of M (which is less than N).

**Approach:**

Following observations need to be made while solving the problem:

- An odd number represented in its binary form has 1 as its Least Significant Bit(LSB), whereas an even number represented in its binary form has 0 as its Least Significant Bit(LSB).
- While performing
**XOR**operation, if the number of set bits is odd, then the XOR value will be 1. If the number of set bits is even, then the XOR value will be 0. - Hence, we need to perform XOR of two odd numbers or two even numbers to obtain an even number as the result.

From the above explanation, the problem can be solved by the following steps:

- If N is odd, the largest odd number which is less than N will be
**N – 2**. - If N is even, the largest even number which is less than N will be
**N – 2**. - Hence, if N = 1, print -1 as a suitable M cannot be obtained in this case. For all other cases, print
**N – 2**as the aswer.

Below is the implementation of the above approach:

## C++

`// C++ program for the above problem. ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to find the maximum ` `// possible value of M ` `int` `getM(` `int` `n) ` `{ ` ` ` `// Edge case ` ` ` `if` `(n == 1) ` ` ` `return` `-1; ` ` ` ` ` `// M = N - 2 is maximum ` ` ` `// possible value ` ` ` `else` ` ` `return` `n - 2; ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` `int` `n = 10; ` ` ` ` ` `int` `ans = getM(n); ` ` ` `cout << ans; ` `} ` |

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## Java

`// Java program for the above problem. ` `import` `java.util.*; ` ` ` `class` `GFG{ ` ` ` `// Function to find the maximum ` `// possible value of M ` `static` `int` `getM(` `int` `n) ` `{ ` ` ` ` ` `// Edge case ` ` ` `if` `(n == ` `1` `) ` ` ` `return` `-` `1` `; ` ` ` ` ` `// M = N - 2 is maximum ` ` ` `// possible value ` ` ` `else` ` ` `return` `n - ` `2` `; ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ` `int` `n = ` `10` `; ` ` ` ` ` `System.out.print(getM(n)); ` `} ` `} ` ` ` `// This code is contributed by sanjoy_62 ` |

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## Python3

`# Python3 program for the above problem ` ` ` `# Function to find the maximum ` `# possible value of M ` `def` `getM(n): ` ` ` ` ` `# Edge case ` ` ` `if` `(n ` `=` `=` `1` `): ` ` ` `return` `-` `1` `; ` ` ` ` ` `# M = N - 2 is maximum ` ` ` `# possible value ` ` ` `else` `: ` ` ` `return` `n ` `-` `2` `; ` ` ` `# Driver code ` `n ` `=` `10` ` ` `print` `(getM(n)) ` ` ` `# This code is contributed by sanjoy_62 ` |

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## C#

`// C# program for the above problem. ` `using` `System; ` `class` `GFG{ ` ` ` `// Function to find the maximum ` `// possible value of M ` `static` `int` `getM(` `int` `n) ` `{ ` ` ` ` ` `// Edge case ` ` ` `if` `(n == 1) ` ` ` `return` `-1; ` ` ` ` ` `// M = N - 2 is maximum ` ` ` `// possible value ` ` ` `else` ` ` `return` `n - 2; ` `} ` ` ` `// Driver code ` `static` `void` `Main() ` `{ ` ` ` `int` `n = 10; ` ` ` ` ` `Console.Write(getM(n)); ` `} ` `} ` ` ` `// This code is contributed by divyeshrabadiya07 ` |

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**Output:**

8

**Time Complexity:** O(1)

**Auxiliary Space:** O(1)