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.util.*;

public class ReversingEncryption {
	static final Input in = new Input();
	static final PrintWriter out = new PrintWriter(System.out);

	public static void main(String[] args) {
		int n=in.nextInt();
		char[] t=in.next().toCharArray();
		for(int i=2; i<=n; i++) {
			if(n%i==0)
				for(int j=0; j<i/2; j++) {
					char temp=t[j];
					t[j]=t[i-j-1];
					t[i-j-1]=temp;
				}
		}
		for(int i=0; i<t.length-1; i++)
			out.print(t[i]);
		out.println(t[t.length-1]);
		out.close();
	}

	static class Input {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st;

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

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