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.IOException;
import java.util.LinkedList;
import java.util.List;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) throws IOException {

        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();
        int k = sc.nextInt();

        String s = sc.next();

        int overlap = 0;

        for(int i = 1; i < n; i ++){
            boolean same = true;
            for(int j = i; j < n; j ++){
                same &= s.charAt(j - i) == s.charAt(j);
                if(!same) break;
            }
            if(same){
                overlap = n - i;
                break;
            }
        }

        String add = s.substring(overlap, n);
        for(int i = 1; i < k; i ++){
            s += add;
        }
        System.out.println(s);
    }
}