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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Harshal Khodifad
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskA solver = new TaskA();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskA {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int n = in.nextInt();
            int k = in.nextInt();

            StringBuilder s = new StringBuilder(in.next());

            StringBuilder t = new StringBuilder(s);

            int l = 1;
            boolean flag = false;

            outer:
            for (l = 1; l <= n; l++) {
                s = new StringBuilder(t);
                s.append(s.substring(n - l, n));

                int x = s.lastIndexOf(t.toString());
                if (x == s.length() - n) {
                    flag = true;
                    break;
                }

//            for (int i=1;i<=s.length()-n;i++)
//            {
//                for (int j=i;j<i+n;j++)
//                {
//                    if (s.charAt(j)!=t.charAt(j-i)){
//                        flag = false;
//                    }
//                }
//            }
            }

            StringBuilder o = new StringBuilder(t);

            if (k == 1) {
                out.print(o);
                return;
            } else {
                while (k-- > 1) {
                    o.append(t.substring(n - l, n));
                }
                out.print(o);
                return;
            }

        }

    }
}