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.Scanner;

public class B999_Reversing_Encryption {
	
	public static int[] divisors;
	public static int index = 0;

	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);
		
		int value = input.nextInt();
		divisors = new int[value + 1];
		divisorFinder(value);
		input.nextLine();
		
		String str = input.nextLine();
		char[] toChar = str.toCharArray();
		
		char temp;
		for(int i = 0; i <= index; i++){
			for(int j = 0, k = divisors[i]-1; j <= (divisors[i]-1)/2; j++, k--){
				temp = toChar[j];
				toChar[j] = toChar[k];
				toChar[k] = temp;
			}
		}
		
		System.out.println(toChar);
		
	}
	
	public static void divisorFinder(int value){
		for(int i = 1; i <= value; i++)
			if(value % i == 0)
				divisors[index++] = i;
		index--;
	}

}
