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.*;

public class Tree {

    public static void main(String[] args) {
        
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        String s = in.next();
        ArrayList<Integer> a = new ArrayList();
        func(a, n);
        Collections.sort(a);
        for(int i=0; i<a.size(); i++)
        {
            s = rev(s, 0, a.get(i)-1);
        }
        System.out.println(s);
    }
    static String rev(String str, int start, int end)
    {
        String ans = "";
        for(int i=end; i>=start; i--)
        {
            ans += str.substring(i, i+1);
        }
        ans += str.substring(end+1);
        return ans;
    }
    static void func(ArrayList<Integer> a, int n)
    {
        for(int i=1; i*i<=n; i++)
        {
            if(n%i == 0)
            {
                if(i*i == n)
                {
                    a.add(i);
                }
                else
                {
                    a.add(i);
                    a.add(n/i);
                }
            }
            
        }
    }
}
