2019寒假培训第一天

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.

代码：

#include<bits/stdc++.h>
using namespace std;
string s;
int a[101];
int main()
{
    int n;
   cin>>n;
   cin>>s;
   int i=n;
   int j=1;
   a[0]=n;
   for(int i=n-1;i>=2;i--)
   {
       if(n%i==0)
        a[j++]=i;

   }
   string s1;
for(int k=j;k>=0;k--)
{
    s1=s.substr(0,a[k]);
    reverse(s1.begin(),s1.end());
    s.replace(0,a[k],s1);
}
  cout<<s<<endl;
 return 0;
}