#include <bits/stdc++.h>
int dx[] = {-1, -1, -1, 0, 1, 1, 1, 0};
int dy[] = {-1, 0, 1, 1, 1, 0, -1, -1};
using namespace std;
long long power(long long a, long long p, long long M) {
  long long r = 1;
  a = a % M;
  while (p > 0) {
    if (p % 2) r = (r * a) % M;
    a = (a * a) % M;
    p /= 2;
  }
  return r % M;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long multiply(long long a, long long b, long long m) {
  return ((a % m) * (b % m)) % m;
}
void solve() {
  long long n;
  scanf("%lld", &n);
  string s;
  cin >> s;
  vector<long long> v;
  for (long long 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());
  for (auto i : v) {
    reverse(s.begin(), s.begin() + i);
  }
  cout << s;
}
int main() {
  solve();
  return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
int dx[] = {-1, -1, -1, 0, 1, 1, 1, 0};
int dy[] = {-1, 0, 1, 1, 1, 0, -1, -1};
using namespace std;
long long power(long long a, long long p, long long M) {
  long long r = 1;
  a = a % M;
  while (p > 0) {
    if (p % 2) r = (r * a) % M;
    a = (a * a) % M;
    p /= 2;
  }
  return r % M;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long multiply(long long a, long long b, long long m) {
  return ((a % m) * (b % m)) % m;
}
void solve() {
  long long n;
  scanf("%lld", &n);
  string s;
  cin >> s;
  vector<long long> v;
  for (long long 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());
  for (auto i : v) {
    reverse(s.begin(), s.begin() + i);
  }
  cout << s;
}
int main() {
  solve();
  return 0;
}

```