#include <bits/stdc++.h>
int main() {
  char s[100];
  char temp;
  int i, j, n;
  scanf("%d", &n);
  scanf("%s", s);
  if (n == 2) {
    temp = s[0];
    s[0] = s[1];
    s[1] = temp;
  } else {
    for (i = 0; i <= n / 2; ++i) {
      if (n % (i + 1) == 0) {
        for (j = 0; j <= i / 2; ++j) {
          temp = s[j];
          s[j] = s[i - j];
          s[i - j] = temp;
        }
      }
    }
    for (i = 0; i < n / 2; ++i) {
      temp = s[i];
      s[i] = s[n - 1 - i];
      s[n - 1 - i] = temp;
    }
  }
  printf("%s", s);
}

### Prompt

Construct a cpp code solution to the problem outlined:

A string s of length n can be encrypted by the following algorithm:

  * iterate over all divisors of n in decreasing order (i.e. from n to 1), 
  * for each divisor d, reverse the substring s[1 ... d] (i.e. the substring which starts at position 1 and ends at position d). 



For example, the above algorithm applied to the string s="codeforces" leads to the following changes: "codeforces" → "secrofedoc" → "orcesfedoc" → "rocesfedoc" → "rocesfedoc" (obviously, the last reverse operation doesn't change the string because d=1).

You are given the encrypted string t. Your task is to decrypt this string, i.e., to find a string s such that the above algorithm results in string t. It can be proven that this string s always exists and is unique.

Input

The first line of input consists of a single integer n (1 ≤ n ≤ 100) — the length of the string t. The second line of input consists of the string t. The length of t is n, and it consists only of lowercase Latin letters.

Output

Print a string s such that the above algorithm results in t.

Examples

Input

10
rocesfedoc


Output

codeforces


Input

16
plmaetwoxesisiht


Output

thisisexampletwo


Input

1
z


Output

z

Note

The first example is described in the problem statement.

### Response

```cpp
#include <bits/stdc++.h>
int main() {
  char s[100];
  char temp;
  int i, j, n;
  scanf("%d", &n);
  scanf("%s", s);
  if (n == 2) {
    temp = s[0];
    s[0] = s[1];
    s[1] = temp;
  } else {
    for (i = 0; i <= n / 2; ++i) {
      if (n % (i + 1) == 0) {
        for (j = 0; j <= i / 2; ++j) {
          temp = s[j];
          s[j] = s[i - j];
          s[i - j] = temp;
        }
      }
    }
    for (i = 0; i < n / 2; ++i) {
      temp = s[i];
      s[i] = s[n - 1 - i];
      s[n - 1 - i] = temp;
    }
  }
  printf("%s", s);
}

```