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.*;
public class stringrev
{
    String reverse(String s)
    {
        int l=s.length();
        String a="";
        for(int i=l-1;i>=0;i--)
    {
        a=a+s.charAt(i);
    }
    return(a);
}
    
public static void main(String args[])throws IOException
{
int l=0;
int i=0;
BufferedReader ea=new BufferedReader(new InputStreamReader(System.in));

l=Integer.parseInt(ea.readLine());


String s=ea.readLine();

stringrev ob=new stringrev();
String st,st2="";
for(i=1;i<l;i++)
{
    if(l%(i+1)==0)
    {
        st=s.substring(0,i+1);
    st2= ob.reverse(st);
    String s3=st2+s.substring(i+1,l);
    s=s3;
}
}
System.out.println(s);
}
}