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.util.*;


public class JavaApplication8 {

   static boolean check(String s,int index){
       int n = s.length() ;
       
       boolean flag = true ;
       
       for(int i = index;i<n;i++){
           if(s.charAt(i) - s.charAt(i-index) != 0){
               flag = false ;
               break ;
           }
       }
       
       return flag ;
   }
   
   public static void main(String[] args) {
       
        Scanner in = new Scanner(System.in) ;
        int n = in.nextInt() ;
        int k = in.nextInt() ;
        String s = in.next() ;
        int l = s.length() ;
        String ans = "" ;
               
        for(int i=1;i<l;i++){
            if(check(s, i)){
                String seg = s.substring(0, i) ;
                
                for(int j=1;j<=k-1;j++){
                    ans += seg ;
                }
                
                ans += s ;
                break ;
            }
        }
        
        if(ans.compareTo("")==0){
            for(int i=1;i<=k;i++){
                ans += s ;
            }
        }
        
        System.out.println(ans) ;
    }
}