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.*;
import java.math.BigInteger;
import java.util.*;

public class Main {
    public static void main(String[] args) throws IOException {
        Locale.setDefault(Locale.US);
        br = new BufferedReader(new InputStreamReader(System.in));
       PrintWriter pw=  new PrintWriter(System.out);
       int n = nextInt();
       String s = next();
       ArrayList<Integer> d = new ArrayList<>();
        for (int i = 1; i < Math.sqrt(n); i++) {
            if (n % i == 0) {
                d.add(i);
                d.add(n / i);
            }
        }
        if ((int)Math.sqrt(n) * (int)Math.sqrt(n) == n) {
            d.add((int) Math.sqrt(n));
        }
        Collections.sort(d);
        char h[] = s.toCharArray();
        for (int i = 0; i < d.size(); i++) {
            char g[] = new char[d.get(i)];
            for (int j = 0; j < d.get(i); j++) {
                g[d.get(i) - j - 1] = h[j];
            }
            for (int j = 0; j < d.get(i); j++) {
                h[j] = g[j];
            }
        }
        for (int i = 0; i < h.length; i++) {
            pw.print(h[i]);
        }
       pw.close();
    }
    static BufferedReader br;
    static StringTokenizer st= new StringTokenizer("");
    public static int nextInt() throws IOException {
        if (!st.hasMoreTokens()) {
            st = new StringTokenizer(br.readLine());
        }
        return Integer.parseInt(st.nextToken());
    }
    public static String next() throws IOException {
        if (!st.hasMoreTokens()) {
            st = new StringTokenizer(br.readLine());
        }
        return st.nextToken();
    }
}