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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Reversing_Encryption {
    public static void main(String[] args) throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int n=Integer.parseInt(br.readLine());
       String b=br.readLine();
       String ans= new String();
       if(b.length()>1) {
           for (int i = 2; i <= n; i++) {
               String x = "";
               if (n % i == 0) {
                   String result = new StringBuilder(b.substring(0, i)).reverse().toString(); //reversed part
                   result += b.substring(i); //after reversing
                   x = result;
                   b = result;
               }

               if (i == n)
                   ans = x;
           }
           System.out.println(ans);
       }
       else System.out.println(b);
        }

}
