#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-9;
const int INFMEM = 63;
const int INF = 1061109567;
const long long LINF = 4557430888798830399LL;
const double DINF = numeric_limits<double>::infinity();
const long long MOD = 1000000007;
const int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1};
const int dy[8] = {0, 1, 0, -1, 1, -1, 1, -1};
const double PI = 3.141592653589793;
inline void fastin(int &input_number) {
  input_number = 0;
  int ch = getchar_unlocked();
  int sign = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') sign = -1;
    ch = getchar_unlocked();
  }
  while (ch >= '0' && ch <= '9') {
    input_number = (input_number << 3) + (input_number << 1) + ch - '0';
    ch = getchar_unlocked();
  }
  input_number *= sign;
}
inline void fastll(long long &input_number) {
  input_number = 0;
  int ch = getchar_unlocked();
  int sign = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') sign = -1;
    ch = getchar_unlocked();
  }
  while (ch >= '0' && ch <= '9') {
    input_number = (input_number << 3) + (input_number << 1) + ch - '0';
    ch = getchar_unlocked();
  }
  input_number *= sign;
}
inline void open(string a) {
  freopen((a + ".in").c_str(), "r", stdin);
  freopen((a + ".out").c_str(), "w", stdout);
}
inline void fasterios() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
string s;
long long n;
int main() {
  cin >> n;
  cin >> s;
  for (int i = 1; i <= n; i++) {
    if (n % i == 0) reverse(s.begin(), s.begin() + i);
  }
  cout << s << '\n';
  return 0;
}

### Prompt

Please formulate 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>
using namespace std;
const double eps = 1e-9;
const int INFMEM = 63;
const int INF = 1061109567;
const long long LINF = 4557430888798830399LL;
const double DINF = numeric_limits<double>::infinity();
const long long MOD = 1000000007;
const int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1};
const int dy[8] = {0, 1, 0, -1, 1, -1, 1, -1};
const double PI = 3.141592653589793;
inline void fastin(int &input_number) {
  input_number = 0;
  int ch = getchar_unlocked();
  int sign = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') sign = -1;
    ch = getchar_unlocked();
  }
  while (ch >= '0' && ch <= '9') {
    input_number = (input_number << 3) + (input_number << 1) + ch - '0';
    ch = getchar_unlocked();
  }
  input_number *= sign;
}
inline void fastll(long long &input_number) {
  input_number = 0;
  int ch = getchar_unlocked();
  int sign = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') sign = -1;
    ch = getchar_unlocked();
  }
  while (ch >= '0' && ch <= '9') {
    input_number = (input_number << 3) + (input_number << 1) + ch - '0';
    ch = getchar_unlocked();
  }
  input_number *= sign;
}
inline void open(string a) {
  freopen((a + ".in").c_str(), "r", stdin);
  freopen((a + ".out").c_str(), "w", stdout);
}
inline void fasterios() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
string s;
long long n;
int main() {
  cin >> n;
  cin >> s;
  for (int i = 1; i <= n; i++) {
    if (n % i == 0) reverse(s.begin(), s.begin() + i);
  }
  cout << s << '\n';
  return 0;
}

```