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.
/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;
import java.util.Arrays;
import java.math.BigInteger;
 
/* Name of the class has to be "Main" only if the class is public. */
public class Main
{
     public static void main (String[] args) throws java.lang.Exception
     {
     	Scanner in =new Scanner(System.in);
		int l,n,k,r,j,t=1,i,m,p,h;
		String s="",st="";
		int x,y,z;
	     for(i=0;i<t;i++)
		{
		     p=h=0;
		     n=in.nextInt();
		     s=in.next();
		     char c[]=s.toCharArray();
		     for(j=2;j<=n;j++)
		     {
		          if(n%j==0)
		          {
		               char ch[]=new char[j];
		               for(k=j-1,l=0;l<j;l++,k--)
		                    ch[l]=c[k];
		               for(l=0;l<j;l++)
		                    c[l]=ch[l];
		          }
		     }
		     System.out.println(String.valueOf(c));
		}
     }
}