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:
//Many Equal Substrings
import java.io.*;
import java.util.*;

public class A506{
   
   public static void main(String[] args)throws IOException{
      BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
      PrintWriter out = new PrintWriter(System.out);
      
      StringTokenizer st = new StringTokenizer(f.readLine());
      
      int n = Integer.parseInt(st.nextToken());
      int m = Integer.parseInt(st.nextToken());
      
      String s = f.readLine();
      
      int index = 0;
      for(int k = 1; k < n; k++){
         //System.out.println(s.substring(k) + " " + s.substring(0,s.length()-k));
         if(s.substring(k).equals(s.substring(0,s.length()-k))){
            index = k;
            break;
         }
      }
      
      String answer = "";
      answer+=s;
      
      for(int k = 0; k < m-1; k++){
         if(index==0) answer+=s;
         else answer+=s.substring(s.length()-index);
      }
      
      out.println(answer);
      
      out.close();
   }
   
}