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;
long long int binaryExponentiation(long long int x, long long int n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return binaryExponentiation(x * x, n / 2);
  else
    return x * binaryExponentiation(x * x, (n - 1) / 2);
}
long long int modularExponentiation(long long int x, long long int n,
                                    long long int M) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return modularExponentiation((x * x) % M, n / 2, M);
  else
    return (x * modularExponentiation((x * x) % M, (n - 1) / 2, M)) % M;
}
bool prime_check(long long x) {
  bool prime = (x >= 2);
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      prime = false;
      break;
    }
  }
  return prime;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  char c;
  cin >> n;
  string s;
  cin >> s;
  vector<int> v;
  for (long long int i = 1; i < n + 1; i++) {
    if (n % i == 0) {
      v.push_back(i);
    }
  }
  for (long long int i = 0; i < v.size(); i++) {
    for (long long int j = 0; j < v[i] / 2; j++) {
      c = s[j];
      s[j] = s[v[i] - 1 - j];
      s[v[i] - 1 - j] = c;
    }
  }
  cout << s;
  return 0;
}
