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.*;
import java.io.*;
public class B
{
	
	
	/********* SOLUTION STARTS HERE ************/
	
	
	
	private static void solve(FastScanner sc, PrintWriter out)
	{
		int n = sc.nextInt();
		String s = sc.next();
		int i = 1;
		ArrayList<Integer> arr = new ArrayList<Integer>();
		for(i = 1 ; i * i <= n ; i++)
		{
			if(n % i == 0)
			{
				arr.add(i);
				if(i != (n / i))	arr.add(n / i);
			}
		}
		Collections.sort(arr);
		StringBuilder sb = new StringBuilder(s);
		for(i = 1 ; i < arr.size() ; i++)
		{
			StringBuilder temp = new StringBuilder(sb.substring(0, arr.get(i)));
			temp.reverse();
			temp.append(sb.substring(arr.get(i)));
			sb = temp;
		}
		System.out.println(sb.toString());
	}
	
	
	
	/************* SOLUTION ENDS HERE **********/
	
	
	
	
	
	/********** TEMPLATE STARTS HERE ***********/
	
	public static void main(String []args) throws IOException 
	{
		FastScanner in  = new FastScanner(System.in);
		PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)), false); 
		solve(in, out);
		in.close();
		out.close();
	}    
	
	static class FastScanner
	{
		BufferedReader reader;
		StringTokenizer st;
		FastScanner(InputStream stream){reader=new BufferedReader(new InputStreamReader(stream));st=null;}	
		String next()
		{while(st == null || !st.hasMoreTokens()){try{String line = reader.readLine();if(line == null){return null;}		    
		st = new StringTokenizer(line);}catch (Exception e){throw new RuntimeException();}}return st.nextToken();}
		String nextLine()  {String s=null;try{s=reader.readLine();}catch(IOException e){e.printStackTrace();}return s;}	    	  	
		int    nextInt()   {return Integer.parseInt(next());}
		long   nextLong()  {return Long.parseLong(next());}		
		double nextDouble(){return Double.parseDouble(next());}
		char   nextChar()  {return next().charAt(0);}
		int[]  nextIntArray(int n)         {int[] arr= new int[n];   int i=0;while(i<n){arr[i++]=nextInt();}  return arr;}
		long[] nextLongArray(int n)        {long[]arr= new long[n];  int i=0;while(i<n){arr[i++]=nextLong();} return arr;}	
		int[]  nextIntArrayOneBased(int n) {int[] arr= new int[n+1]; int i=1;while(i<=n){arr[i++]=nextInt();} return arr;}	    	
		long[] nextLongArrayOneBased(int n){long[]arr= new long[n+1];int i=1;while(i<=n){arr[i++]=nextLong();}return arr;}	    	
		void   close(){try{reader.close();}catch(IOException e){e.printStackTrace();}}				
	}
	
	/*********** TEMPLATE ENDS HERE *************/
}
