#include <bits/stdc++.h>
using namespace std;
vector<int> fac;
void factor(int n, int m) {
  if (m * m == n) {
    fac.push_back(m);
    return;
  } else if (m * m > n) {
    return;
  }
  if (n % m == 0) {
    fac.push_back(m);
    factor(n, m + 1);
    fac.push_back(n / m);
  } else {
    factor(n, m + 1);
  }
}
int main() {
  int n;
  cin >> n;
  string str;
  cin >> str;
  factor(n, 1);
  for (vector<int>::iterator it = fac.begin(); it != fac.end(); it++) {
    reverse(str.begin(), str.begin() + *it);
  }
  cout << str << endl;
}

### Prompt

Create a solution in CPP for the following problem:

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>
using namespace std;
vector<int> fac;
void factor(int n, int m) {
  if (m * m == n) {
    fac.push_back(m);
    return;
  } else if (m * m > n) {
    return;
  }
  if (n % m == 0) {
    fac.push_back(m);
    factor(n, m + 1);
    fac.push_back(n / m);
  } else {
    factor(n, m + 1);
  }
}
int main() {
  int n;
  cin >> n;
  string str;
  cin >> str;
  factor(n, 1);
  for (vector<int>::iterator it = fac.begin(); it != fac.end(); it++) {
    reverse(str.begin(), str.begin() + *it);
  }
  cout << str << endl;
}

```