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.
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    
    public static void main(String[] args) {
   
    Scanner in = new Scanner(System.in);
    
        int n = in.nextInt();
        String s = in.next();
        
        ArrayList<Integer> al = new ArrayList<>();
        for(int i=1; i<=n; i++)
        {
            if(n%i==0)
            {
                al.add(i);
            }
        }
        
        char a[] = s.toCharArray();
        
        for(int i=0; i<al.size(); i++)
        {
            int d = al.get(i)-1;
            int l = 0;
            while(l<d)
            {
                char temp = a[l];
                a[l] = a[d];
                a[d] = temp;
                l++;
                d--;
            }
        }
        
        System.out.println(String.valueOf(a));
        
    }      
}