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;

/**
 * @author Wchert
 */
public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int k = in.nextInt();
        String pattern = in.next();
        char[] p = pattern.toCharArray();
        int[] next = next(p);
//        StringBuilder sb = new StringBuilder(n * k - (k - 1) * (n - next[n] - 1));//空间复杂度优化
        System.out.print(pattern);
        String end = pattern.substring(next[n]);
        while (--k > 0) {
            System.out.print(end);
            /*for (int i = next[n]; i < n; i++) {
                sb.append(p[i]);
            }*/
        }
        System.out.println();
    }

    private static int[] next(char[] p) {
        int len = p.length;
        int[] next = new int[len + 1];
        int i = 0;
        int n = next[0] = -1;
        while (i < len) {
            if (n == -1 || p[n] == p[i]) {
                next[++i] = ++n;
            } else {
                n = next[n];
            }
        }
        return next;
    }
}