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.*;
import java.util.*;
public class Solution
{
    public static void main(String[] args) throws IOException
    {
        BufferedReader ob=new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(ob.readLine());
        int a[]=new int[n];
        int k=0;
        String s=ob.readLine();
        for (int i=1; i<=Math.sqrt(n); i++)
        {
            if (n%i==0)
            {
                // If divisors are equal, print only one
                if (n/i == i)
                {
                    a[k]=i; k++;
                }

                else // Otherwise print both
                {
                    a[k]=i; k++; a[k]=n/i; k++;
                }
            }
        }
        Arrays.sort(a);
        for(int i=0;i<n;i++)
        {
            if(a[i]!=0 && a[i]!=1)
            {
                String h=s.substring(0,0+a[i]);
                char st[]=h.toCharArray();
                String st2="";
                for(int j=h.length()-1;j>=0;j--)
                {
                    st2=st2+st[j];
                }
                s=st2+s.substring(0+a[i],n);
            }
        }
        System.out.println(s);
    }

}