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.BufferedReader;
import java.io.InputStreamReader;
import java.util.Scanner;

/**
 *
 * @author Artemon
 */
public class A {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n, k;
        String t;

        n = in.nextInt();
        k = in.nextInt();
        t = in.next();
        int b = 5;

        int pil = findPiLength(t);

        StringBuilder sb = new StringBuilder();
        String s = t.substring(pil);
        sb.append(t);

        for (int i = 1; i < k; i++) {
            sb.append(s);
        }
        System.out.print(sb);
    }

    public static int findPiLength(String t) {
        char[] cht = t.toCharArray();
        int res = 0;
        int lci, rci;
        int length = cht.length;
        String s = t.substring(0, length - 1);

        rci = length - 1;
        lci = s.lastIndexOf(cht[length - 1]);

        while (true) {
            int lcci = lci;
            while (lci >= 0 && (cht[lci] == cht[rci])) {
                lci--;
                rci--;
                res++;
            }
            if (lci == -1) {
                break;
            } else {
                res = 0;
                s = t.substring(0, lcci);
                lci = s.lastIndexOf(cht[length - 1]);
                rci = length - 1;
            }
        }
        return res;
    }
}