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.
#include <bits/stdc++.h>
int dr[] = {2, 2, -2, -2, 1, -1, 1, -1};
int dc[] = {1, -1, 1, -1, 2, 2, -2, -2};
int dr1[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dc1[] = {1, -1, 1, 0, -1, 0, 1, -1};
int dr2[] = {0, 0, 1, -1};
int dc2[] = {1, -1, 0, 0};
using namespace std;
int main() {
  long long int n, k, i, j;
  while (cin >> n) {
    vector<long long int> v;
    string t;
    cin >> t;
    for (i = 1; i <= sqrt(n); i++) {
      if (n % i == 0) {
        v.push_back(i);
        if (i * i != n) v.push_back(n / i);
      }
    }
    sort(v.begin(), v.end());
    for (i = 0; i < v.size(); i++) {
      long long int p = v[i];
      string t1;
      for (j = 0; j < p; j++) {
        t1.push_back(t[j]);
      }
      reverse(t1.begin(), t1.end());
      for (j = 0; j < p; j++) t[j] = t1[j];
    }
    cout << t << endl;
  }
  return 0;
}
