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.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {
 public static void main(String[] args) throws Exception {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));

        
        String[] s1=br.readLine().split(" ");
        int t=Integer.parseInt(s1[0]);
        int k=Integer.parseInt(s1[1]);

        String str=br.readLine();

        int cnt=createLps(str);

        String s=str.substring(0,t-cnt);

        String res="";
        for(int i=0;i<k;i++)res+=s;
        res+=str.substring(t-cnt,t);
        System.out.println(res);
       

    }

    public static int createLps(String str){
        int len=0;
        int i=1;
        int n=str.length();
        int[] lps=new int[n];
        while(i<n){
            if(str.charAt(i)==str.charAt(len)){
                lps[i]=len+1;
                len++;
                i++;
            }else{
                if(len==0){
                    lps[i]=0;
                    i++;
                }else{
                    len=lps[len-1];
                }
            }
        }
        return lps[n-1];
    }

    
}