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.util.Scanner;

public class RevEncryp {

	public static void main(String[] args) {
		
		Scanner s = new Scanner(System.in);
		
		int n = s.nextInt();
		String t = s.next();
		
		char a [] = new char [n];
		int i;
		for(i=0;i<n;i++){
			a[i] = t.charAt(i);	
		}
		
		
		
		for(i=1;i<=n;i++){	
			if(n%i==0){

				a = rev(a,i);
		
				
			}
		}
				
		
		for(i=0;i<n;i++){
			System.out.print(a[i]);
		}
		
	}
	
	public static char[] rev(char [] a ,int k){
		
		int i=0;
		int j=k-1;
		
		while(i<j && i!=j){
			char temp = a[i];
			a[i] = a[j];
			a[j] = temp;
			i++;
			j--;
		}
		return a;
	}

	
}
