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.
/*input
1
z

*/
import java.util.*;
public class Main
{
	public static String reverse(String s, int d)
	{
		// System.out.println(d);
		String str=s.substring(0,d);
		// System.out.println(str);
		String rest="";
		// if(d!=s.length()-1)
			rest=s.substring(d);
		// System.out.println(rest);

		return (rev(str)+rest);
	}

	public static String rev(String s)
	{
		StringBuffer sb=new StringBuffer();
		int n=s.length();
		for(int i=n-1;i>=0;i--)
		{
			sb.append(s.charAt(i));
		}
		return sb.toString();
	}
	public static void main(String[] args) 
	{
		Scanner sc = new Scanner(System.in);
		int n=sc.nextInt();
		String s=sc.next();
		for(int i=1;i<=n;i++)
		{
			if(n%i==0)
			{
				s=reverse(s,i);
				// System.out.println(s);
			}
		}
		System.out.println(s);	
	}
}