Reversing Encryption（模拟水题）

A string ss of length nn can be encrypted（加密） by the following algorithm:

• iterate（迭代） over all divisors（除数） of nn in decreasing order (i.e. from nn to 11),
• for each divisor dd, reverse the substring s[1…d]s[1…d] (i.e. the substring which starts at position 11 and ends at position dd).

For example, the above algorithm applied to the string ss="codeforces" leads to the following changes: "codeforces" →→ "secrofedoc" →→ "orcesfedoc" →→"rocesfedoc" →→ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because d=1d=1).

You are given the encrypted string tt. Your task is to decrypt this string, i.e., to find a string ss such that the above algorithm results in string tt. It can be proven that this string ss always exists and is unique.

Input

The first line of input consists of a single integer nn (1≤n≤1001≤n≤100) — the length of the string tt. The second line of input consists of the string tt. The length of tt is nn, and it consists only of lowercase Latin letters.

Output

Print a string ss such that the above algorithm results in tt.

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.

 

题目意思：对所给的字符串进行加密得到一新的字符串，加密方式是从n的最小因子开始，反转从开始到因子的字符串。

之前做这道题的时候理解错意思了，一直以为是一半一半的反转字符串，很是可惜，没有做出来

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
    int n,i;
    char s[110];
    scanf("%d",&n);
    getchar();
    scanf("%s",&s);
    for(i=1;i<=n;i++)
    {
        if(n%i==0)
        {
            reverse(s,s+i);
        }
    }
    printf("%s\n",s);
    return 0;
}

 这里有我之前一篇关于字符串反转函数的博客

 

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
    int n,i,j,count;
    char s[110];
    int a[110];
    scanf("%d",&n);
    getchar();
    scanf("%s",&s);
    count=0;
    for(i=1;i<=n;i++)
    {
        if(n%i==0)
        {
            a[count++]=i;///记录n因子
        }
    }
    for(i=0;i<count;i++)
    {
        for(j=0;j<a[i]/2;j++)
        {
            swap(s[j],s[a[i]-j-1]);
        }
    }
    printf("%s\n",s);
    return 0;
}