CodeForces - 999B Reversing Encryption

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 tis 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.

            

        题目的意思就是例子里面写的那样，已知这个字符串的长度为n，如果一个数d为它的因子，那么从1到d都要逆序（字符串从1开始计数），将所有的因子求出，依次进行逆序就可以了，但是要注意的是，这里是从最小的因子开始逆序。

#include <bits/stdc++.h>

using namespace std;

char s[105];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        getchar();
        scanf("%s",s);
        for(int i = 1; i < n ; i ++)
        {
            if(n % i == 0)
            {
                for(int j = 0; j < i / 2; j ++)
                {
                    char op = s[j];
                    s[j] = s[i - j - 1];
                    s[i-j -1 ] = op;
                }
            }
        }
        for(int i = 0; i <=(n -1)/2; i ++)
        {
            char op = s[i];
            s[i] = s[n - i - 1];
            s[n - i - 1] = op;
        }
        printf("%s",s);
        printf("\n");
    }
    return 0;
}