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.IOException;
import java.util.LinkedList;
import java.util.List;
import java.util.Scanner;

public class Main {
	public static void main(String args[]) throws IOException {
		int n;
		String s;
		Scanner sc = new Scanner(System.in);
		n = sc.nextInt();
		s = sc.next();
		List<Integer> l = new LinkedList<>();
		// StringBuilder sb = new StringBuilder(s);
		for (int i = 1; i <= n; i++) {
			if (n % i == 0) {
				l.add(i - 1);
			}
		}
		
		StringBuilder sb = new StringBuilder("");
		if(l.size()==1)
		sb.append(s.charAt(0));
		for (int i = 0; i < l.size() - 1; i++) {
			int j=l.get(i)+1;
			StringBuilder k = new StringBuilder("");
			if(i==0)
			j = l.get(i);
			for (; j <= l.get(i + 1); j++) {
				k.append(s.charAt(j));
			}
			//System.out.println(sb+ " "+k);
			sb.append(k);
			sb.reverse();
			
		}
		System.out.println(sb);
	}
}
