#include <bits/stdc++.h>
void swap(char *buf, int k) {
  int i;
  char a;
  for (i = 0; i < k / 2; i++) {
    a = buf[i];
    buf[i] = buf[k - 1 - i];
    buf[k - 1 - i] = a;
  }
}
int main(void) {
  char buf[101];
  int i, n;
  scanf("%d\n", &n);
  if (!fgets(buf, 101, stdin)) {
    printf("Cannot read\n");
    return 0;
  }
  for (i = 0; buf[i]; i++) {
    if (buf[i] == '\n') {
      buf[i] = 0;
      break;
    }
  }
  for (i = 2; i <= n; i++)
    if (n % i == 0) {
      swap(buf, i);
    }
  printf("%s\n", buf);
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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>
void swap(char *buf, int k) {
  int i;
  char a;
  for (i = 0; i < k / 2; i++) {
    a = buf[i];
    buf[i] = buf[k - 1 - i];
    buf[k - 1 - i] = a;
  }
}
int main(void) {
  char buf[101];
  int i, n;
  scanf("%d\n", &n);
  if (!fgets(buf, 101, stdin)) {
    printf("Cannot read\n");
    return 0;
  }
  for (i = 0; buf[i]; i++) {
    if (buf[i] == '\n') {
      buf[i] = 0;
      break;
    }
  }
  for (i = 2; i <= n; i++)
    if (n % i == 0) {
      swap(buf, i);
    }
  printf("%s\n", buf);
  return 0;
}

```