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.
/*
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 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */
// package reversing.encryption;

import java.util.Arrays;
import java.util.Scanner;

/**
 *
 * @author Mostafa
 */
public class ReversingEncryption {

    /**
     * @param args the command line arguments
     */
    
    public static void main(String[] args) {
        // TODO code application logic here
        Scanner in = new Scanner(System.in) ;
        byte n = in.nextByte() ;
        String s = in.next() ;
        char x[] = s.toCharArray() ;
        for ( int i = 2 ; i <= n ; ++i )
        {
            if ( n % i == 0 )
            {
                for ( byte j = 0 ; j*2 <= i-1 ; ++j )
                {
//                    System.out.println(x[j] + " " + x[i-j]);
                    char t = x[j] ;
                    x[j] = x[i-j-1];
                    x[i-j-1] = t ;
//                    System.out.println(x[j] + " " + x[i-j]);
//                    System.out.println();
                }
            }
        }
        System.out.println(x);
    }
    
}
