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 B {
    public static String reverseString(String str){  
        StringBuilder sb=new StringBuilder(str);  
        sb.reverse();  
        return sb.toString();  
    }  
    static void swap(char a, char b){
        char temp = a;
        a = b;
        b = temp;
    }
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int t = 1;
        while(t-->0){
            int n = sc.nextInt();
            String str = sc.next();
            char[] ch = str.toCharArray();
            for(int i=1; i<=n; i++){
                if(n%i==0){
                    int start = 0;
                    int end = i-1;
                    while(start<=end){
                        // swap(ch[start], ch[end]);
                        char temp = ch[start];
                        ch[start] = ch[end];
                        ch[end] = temp;
                        start++;
                        end--;
                    }
                }
            }
            System.out.println(String.valueOf(ch));
        }
        sc.close();
    }
}