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.
/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */
import java.util.*;
/**
 *
 * @author paras
 */
public class div3 {
    public static void main(String arg[])
    {
 Scanner ob=new Scanner(System.in);   
 int n=ob.nextInt();
 String s=ob.next();
 StringBuilder ab=new StringBuilder();
 StringBuilder ab1=new StringBuilder();

 //ab.append(s);
 for(int i=0;i<n;i++)
 {int y=i+1;
 ab.append(s.charAt(i));
     if(n%(y)==0)
     {
         ab=ab.reverse();
      }
      
 }ab1=ab1.reverse();
        System.out.println(ab.toString());
    }
}
