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.*;
/**
 *
 * @author comp
 */
public class NewClass {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        String s = in.next();
        StringBuffer sb = new StringBuffer(s);
        ArrayList a = new ArrayList();
        for (int i = 1; i <= n; i++) {
            if ((n % i) == 0){
                a.add(i);
            }
        }
        Integer f[] = (Integer[])a.toArray(new Integer[a.size()]);
        for (int i = 0; i < f.length; i++) {
            for (int j = 0; j < f[i]/2; j++) {
                char c = sb.charAt(j);
                sb.setCharAt(j, sb.charAt(f[i]-j-1));
                sb.setCharAt((f[i]-j-1), c);
                
            }
        }
        System.out.println(sb);
    }
}