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.List;
import java.util.Scanner;

public class Main {

   public static void main(String[] args) {
       Scanner in = new Scanner(System.in);

       int n;
       String s;

       n = in.nextInt();
       s = in.next();

       List<Integer> divisors = divisors(n);

       for (Integer divisor : divisors) {
            s = reverse(s, divisor);
       }

       System.out.println(s);
   }

   private static String reverse(String s, Integer divisor) {
       String rev = s.substring(0, divisor);
       char [] a = new char[rev.length()];
       for(int i = 0; i<rev.length();i++) {
           a[i] = (rev.charAt(rev.length() - i - 1));
       }

       return new String(a) + s.substring(divisor, s.length());
   }

   private static List<Integer> divisors(int n) {
       List<Integer> ans = new ArrayList<>();
       for(int i = 2; i<=n;i++) {
           if(n%i == 0) {
               ans.add(i);
           }
       }

       return ans;
   }
}