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.*;
import java.io.*;

public class Equal{
    
    static int N, M, K;
    static String s;
    static StringTokenizer st;
    static int[] d;
    
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));        
        PrintWriter out = new PrintWriter(System.out);
        
        
        st = new StringTokenizer(br.readLine());
        int N = Integer.parseInt(st.nextToken());
        int K = Integer.parseInt(st.nextToken());
        
        String s  = br.readLine();
        
        
        out:
        for (int i = N-1; i >= 0; i--) {
            for (int j = N-i; j < N; j++) {
                if(s.charAt(j-N+i) != s.charAt(j)){
                    continue out;
                }
            }
            out.print(s);
            String ad = s.substring(i);
            while (--K > 0) {
                out.print(ad);
            }
            out.println();
            break;
        }
        out.close();
    }    
}