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.Scanner;

public class Main {
	
	private static boolean check(String s , int pos , String target) {
		
		int i = pos , j = 0;
		while (i < s.length() && j < target.length() && (s.charAt(i) == target.charAt(j))) {
			i ++;
			j ++;
		}
		if (i == s.length()) {
			return true;
		} else {
			return false;
		}
		
	}

	public static void main(String[] args) throws Exception {
		
		Scanner scan = new Scanner(System.in);

		int i , j = 1 , n , k;
		n = scan.nextInt();
		k = scan.nextInt();
		
		String string = scan.next();
		String s = string;
	
		for (i = 0;i < k - 1;i ++) {
			while (j < s.length() && check(s , j , string) == false) {				
				j ++;
			}
			if (j == s.length()) {
				s += string;
				j += 1;
			} else {
				int index = 0 , temp = j;
				while (j < s.length() && index < string.length()) {
					j ++;
					index ++;
				}
				if (index == string.length()) {
					j = temp + 1;
				} else {
					s += string.substring(index);
					j = temp + 1;
				}
			}
		}
		System.out.println(s);
		
	}
	
    
}