Problem: You are given a string t consisting of n lowercase Latin letters and an integer number k.

Let's define a substring of some string s with indices from l to r as s[l ... r].

Your task is to construct such string s of minimum possible length that there are exactly k positions i such that s[i ... i + n - 1] = t. In other words, your task is to construct such string s of minimum possible length that there are exactly k substrings of s equal to t.

It is guaranteed that the answer is always unique.

Input

The first line of the input contains two integers n and k (1 ≤ n, k ≤ 50) — the length of the string t and the number of substrings.

The second line of the input contains the string t consisting of exactly n lowercase Latin letters.

Output

Print such string s of minimum possible length that there are exactly k substrings of s equal to t.

It is guaranteed that the answer is always unique.

Examples

Input

3 4
aba


Output

ababababa


Input

3 2
cat


Output

catcat

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.lang.*;
import java.awt.*;
import java.awt.geom.*;
import java.math.*;
import java.text.*;
public class   cf506p1
{
  BufferedReader in;
  StringTokenizer as;
  int nums[],nums2[];
  ArrayList < Integer > [] ar;
  
  public static void main (String[] args)
  {
    new cf506p1  ();
  }
  
  public cf506p1  ()
  {
    try
    {
      
      in = new BufferedReader (new InputStreamReader (System.in));
      int a = nextInt();
      int b = nextInt();
      String c = nextLine();
      String out = " ";
      out+=c;
      int check = 1;
      boolean done = false;
      while(check < b)
      {
        int first =a;
        for(int x = out.length()-a+1;x<=out.length();x++)
        {
          first--;

          String temp = out.substring(x);
          if(c.indexOf(temp) == 0)
          {
            break;
          }
        }
        check++;
        out += c.substring(first);
      }
      System.out.println(out);
    }
    
    catch(IOException e)
    {
      
    }
  }
  
  
  
  
  
  
  
  String next () throws IOException
  {
    while (as == null || !as.hasMoreTokens ())
    {
      as = new StringTokenizer (in.readLine ().trim ());
    }
    
    
    return as.nextToken ();
  }
  
  
  
  long nextLong () throws IOException
  {
    return Long.parseLong (next ());
  }
  
  
  int nextInt () throws IOException
  {
    return Integer.parseInt (next ());
  }
  
  
  double nextDouble () throws IOException
  {
    return Double.parseDouble (next ());
  }
  
  
  String nextLine () throws IOException
  {
    return in.readLine ().trim ();
  }
}