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.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = Integer.parseInt(sc.next());
        int k = Integer.parseInt(sc.next());
        String s = sc.next();
        String res = s;
        int l = 0;
        String add = null;
        for(int i=1;i<res.length();i++){
            String ost = res.substring(i, res.length());
            String s1 = s.substring(0, ost.length());
            if(ost.equals(s1)) {
                add = s.substring(ost.length(), s.length());
                break;
            }
        }
        if(add == null) {
            for(int i=1;i<k;i++){
                res+=s;
            }
            System.out.println(res);
            return;
        }
        for(int i=1;i<k;i++){
            res+=add;
        }
        System.out.println(res);
    }
}