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

public class CodeTest {
	static ArrayList<Integer> set = new ArrayList<Integer>();
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n= sc.nextInt();
		String s = sc.next();
		printDivisors(n);
		set.sort(null);
		char c[] = s.toCharArray();
		for(int i = 1; i<set.size();i++) {
			int k = set.get(i);
			for (int left = 0; left < k; left++, k--) {
	            char temp = c[left];
	            c[left] = c[k-1];
	            c[k-1] = temp;
	        }
		}
		for(int i=0;i<c.length;i++)
			System.out.print(c[i]);
	}
	
	
	static void printDivisors(int n) 
    { 
        // Note that this loop runs till square root 
        for (int i=1; i<=Math.sqrt(n); i++) 
        { 
            if (n%i==0) { 
            	if (n/i == i) 
                    set.add(i); 
       
                else {
                	set.add(i);
                	set.add(n/i);
                }
            } 
        } 
    } 
}