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;
import java.util.*;

public class Main {

    public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        int n = sc.ni();
        char[] a = sc.next().toCharArray();
        for (int i = 2; i <= n; i++) {
            if (n % i == 0)
                reverse(a, i);
        }
        System.out.println(new String(a));
    }

    static void reverse(char[] a, int r) {
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < r; i++) {
            sb.append(a[i]);
        }
        sb.reverse();
        for (int i = 0; i < r; i++)
            a[i] = sb.toString().charAt(i);
    }


    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner() {
            br = new BufferedReader(new InputStreamReader(System.in), 32768);
            st = null;
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int ni() {
            return Integer.parseInt(next());
        }

        int[][] graph(int N, int[][] edges) {
            int[][] graph = new int[N][];
            int[] sz = new int[N];
            for (int[] e: edges) {
                sz[e[0]] += 1;
                sz[e[1]] += 1;
            }
            for (int i = 0; i < N; i++) {
                graph[i] = new int[sz[i]];
            }
            int[] cur = new int[N];
            for (int[] e: edges) {
                graph[e[0]][cur[e[0]]] = e[1];
                graph[e[1]][cur[e[1]]] = e[0];
                cur[e[0]] += 1;
                cur[e[1]] += 1;
            }
            return graph;
        }

        int[] intArray(int N, int mod) {
            int[] ret = new int[N];
            for (int i = 0; i < N; i++)
                ret[i] = ni()+mod;
            return ret;
        }

        long nl() {
            return Long.parseLong(next());
        }

        long[] longArray(int N, long mod) {
            long[] ret = new long[N];
            for (int i = 0; i < N; i++)
                ret[i] = nl()+mod;
            return ret;
        }

        double nd() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}
