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;
std::string Round(std::string &str_d) {
  auto iter_dot = find(str_d.begin(), str_d.end(), '.');
  if (iter_dot != str_d.end() && (distance(iter_dot, str_d.end()) > 2)) {
    char ch = *(iter_dot + 2);
    if (ch > '4') {
      if (*(iter_dot + 1) != '9')
        *(iter_dot + 1) += 1;
      else {
        *(iter_dot + 1) = '0';
        if (*(iter_dot - 1) != '9')
          *(iter_dot - 1) += 1;
        else {
          if (std::distance(str_d.begin(), iter_dot) > 1) {
            *(iter_dot - 1) = '0';
            *(iter_dot - 2) += 1;
          } else {
            *(iter_dot - 1) = '0';
            str_d.insert(str_d.begin(), '1');
            iter_dot = find(str_d.begin(), str_d.end(), '.');
          }
        }
      }
    }
    str_d = std::string(str_d.begin(), iter_dot + 2);
  }
  return str_d;
}
string reBar(const string &str, int pos) {
  string str_bar = string(str.begin(), str.begin() + pos);
  reverse(str_bar.begin(), str_bar.end());
  string str_res = str_bar + string(str.begin() + pos, str.end());
  return str_res;
}
int main() {
  cin.sync_with_stdio(false);
  cout.sync_with_stdio(false);
  int n;
  string str;
  cin >> n;
  cin >> str;
  size_t Size = str.size();
  for (int i = 1; i <= Size; ++i) {
    if (Size % i == 0) str = reBar(str, i);
  }
  cout << str << endl;
  return 0;
}
