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 long long mod = 1e9 + 7;
long long expo(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y % 2) res = (res * x % mod) % mod;
    x = (x * x) % mod;
    y /= 2;
  }
  return res;
}
long long inv(long long x) { return expo(x, mod - 2); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  string s;
  cin >> s;
  vector<long long> div;
  for (long long i = 2; i <= n; i++)
    if (n % i == 0) div.push_back(i);
  if (div.size() == 0) {
    cout << s << "\n";
    return 0;
  }
  sort(div.begin(), div.end());
  for (long long i = 0; i < div.size(); i++) {
    string t = "";
    for (long long j = 0; j < div[i]; j++) t += s[j];
    reverse(t.begin(), t.end());
    string t2 = "";
    for (long long j = 0; j < div[i]; j++) t2 += t[j];
    for (long long j = div[i]; j < n; j++) t2 += s[j];
    s = t2;
  }
  cout << s;
  cout << "\n";
}
