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 HelloWorld
{
  // arguments are passed using the text field below this editor
  public static void main(String[] args)
  {
    Scanner in=new Scanner(System.in);
    int n=in.nextInt();
    in.nextLine();
    String t=in.nextLine();
    char[]ar=t.toCharArray();
    for (int i=1;i<=n/2;i++){
        if (n%i==0){
            for(int j=0;j<i/2;j++){
                char c=ar[j];
                ar[j]=ar[i-1-j];
                ar[i-1-j]=c;
            }
        }
    }
    
    for(int j=0;j<n/2;j++){
        char c=ar[j];
        ar[j]=ar[n-1-j];
        ar[n-1-j]=c;
    }
    System.out.println(new String(ar));
  }
}