You are given an even integer n​​​​​​. You initially have a permutation perm of size n​​ where perm[i] == i​ (0-indexed)​​​​.

In one operation, you will create a new array arr, and for each i:

• If i % 2 == 0, then arr[i] = perm[i / 2].
• If i % 2 == 1, then arr[i] = perm[n / 2 + (i - 1) / 2].

You will then assign arr​​​​ to perm.

Return the minimum non-zero number of operations you need to perform on perm to return the permutation to its initial value.

Example 1:

Input: n = 2
Output: 1
Explanation: prem = [0,1] initially.
After the 1st operation, prem = [0,1]
So it takes only 1 operation.


Example 2:

Input: n = 4
Output: 2
Explanation: prem = [0,1,2,3] initially.
After the 1st operation, prem = [0,2,1,3]
After the 2nd operation, prem = [0,1,2,3]
So it takes only 2 operations.


Example 3:

Input: n = 6
Output: 4


Constraints:

• 2 <= n <= 1000
• n​​​​​​ is even.

## Solution: Brute Force / Simulation

Time complexity: O(n2) ?
Space complexity: O(n)

## C++

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