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;
void watchset_e(set<int> SET, int N) {
  set<int>::iterator ITERATOR = SET.begin();
  advance(ITERATOR, N);
  printf("Element in %dth = %d \n", N, *ITERATOR);
}
int SD() {
  int x;
  scanf("%d", &x);
  return x;
}
long long SLL() {
  long long x;
  scanf("%lld", &x);
  return x;
}
char SC() {
  char x;
  scanf("%c", &x);
  return x;
}
const int AKA = 1e6 + 5;
vector<int> prime;
long long notPrime[AKA];
void seive() {
  notPrime[0] = notPrime[1] = 1;
  for (int i = 2; i <= AKA - 5; ++i) {
    if (!notPrime[i]) prime.push_back(i);
    for (long long j = 1LL * i * i; j <= AKA - 5; j += i) notPrime[j] = 1;
  }
}
vector<int> divs;
void genDivs(int n) {
  for (int i = 1; i * i <= n; ++i) {
    if (n % i == 0) {
      divs.push_back(i);
      if (n / i != i) divs.push_back(n / i);
    }
  }
}
int arr[AKA], fr[AKA];
set<string> s;
int main() {
  int ACCEPTED = 1;
  ;
  while (ACCEPTED--) {
    int n = SD();
    genDivs(n);
    sort(divs.begin(), divs.end());
    string s;
    cin >> s;
    int str = s.length(), vs = divs.size();
    for (int i = 0; i < vs; ++i) {
      reverse(s.begin(), s.begin() + divs[i]);
    }
    cout << s;
  }
  return 0;
}
