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.*;

public class serv{
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		String str = in.next();
		char[] ch = str.toCharArray();
		String ans = "";

		//rocesfedoc
		//codefsecor

		for (int i=2;i<=n;i++) {
			if (n % i == 0) {
				str = turn(str, i);
				
			}
		}
		System.out.println(str);
		
	}
	public static String turn(String str, int len){
		String answ = "";
		String sub = str.substring(0, len);
		for (int j=sub.length()-1;j>=0;j--) {
			answ = answ + sub.charAt(j);
		}
		
		String sub1 = str.substring(len, str.length());
		answ = answ + sub1;
		return answ;

	}
}