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.*;
import java.io.*;

public class Main {
    public static void main(String[] args) throws Exception {
        new Solver().run(1);
    }
}

class Solver {

    private BufferedReader reader = null;
    private StringTokenizer st = null;

    private char[] str;
    private int n;

    public void run(int inputType) throws Exception {
        if (inputType == 0)
            reader = new BufferedReader(new FileReader("input.txt"));
        else
            reader = new BufferedReader(new InputStreamReader(System.in));

        n = Integer.parseInt(reader.readLine());
        str = reader.readLine().toCharArray();

        for (int i = 1; i <= n; ++i) {
            if (n % i > 0)
                continue;

            for (int j = 0; j < i / 2; ++j) {
                char tmp = str[j];
                str[j] = str[i - j - 1];
                str[i - j - 1] = tmp;
            }
        }

        System.out.println(String.valueOf(str));

        reader.close();
    }
}
