B - Reversing Encryption

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.

AC代码：（先遍历把倍数找到然后不断进行替换，

很坑一开始看到题目中的解释我以为是从后面进行替换结果是从前面开始）

#include <iostream>
#include <bits/stdc++.h>
#include <string.h>
using namespace std;

int main()
{
    int a[120], i, n, h, j, k;
    char s[120], t;
    while(cin>>n>>s)
    {
     h = 0;
    for(i = 2;i<=n;i++)
    {
     if(n%i==0)
     a[h++] = i;
    }
    for(i = 0;i<h;i++)
    {
     for(j = a[i]-1, k = 0;k<=j;j--,k++)
     {
      t = s[j],s[j] = s[k], s[k] = t;
     }
    }
    for(i = 0;i<n;i++)
    {
     printf("%c",s[i]);
    }
    printf("\n");
    }
    return 0;
}