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>
using namespace std;
const int MOD = 1000000007;
const int MAX = 1000005;
int SetBit(int n, int x) { return n | (1 << x); }
int ClearBit(int n, int x) { return n & ~(1 << x); }
int ToggleBit(int n, int x) { return n ^ (1 << x); }
bool CheckBit(int n, int x) { return (bool)(n & (1 << x)); }
int main(void) {
  int n, i, j, h, k, len;
  string str;
  while (scanf("%d", &n) != EOF) {
    cin >> str;
    vector<int> v;
    for (i = 1; i * i <= n; i++) {
      if (n % i == 0) {
        v.push_back(i);
        if (i * i != n) v.push_back(n / i);
      }
    }
    sort(v.begin(), v.end());
    len = v.size();
    for (i = 0; i < len; i++) {
      k = v[i];
      h = v[i] / 2;
      for (j = 0; j < h; j++) swap(str[j], str[k - 1 - j]);
    }
    cout << str << "\n";
  }
  return 0;
}
