# GeetCode Hub

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: perm = [0,1] initially.
After the 1st operation, perm = [0,1]
So it takes only 1 operation.
```

Example 2:

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

Example 3:

```Input: n = 6
Output: 4
```

Constraints:

• `2 <= n <= 1000`
• `n`​​​​​​ is even.

class Solution { public int reinitializePermutation(int n) { } }