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 BSRK Aditya
 */
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);
        AManyEqualSubstrings solver = new AManyEqualSubstrings();
        solver.solve(1, in, out);
        out.close();
    }

    static class AManyEqualSubstrings {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int n = in.nextInt(), k = in.nextInt();
            String t = in.next();
            int i;
            for (i = 1; i <= n; i++) {
                int len = n - i;
                String tStart = t.substring(0, n - i);
                String tEnd = t.substring(i);
                //out.println("tStart: " + tStart);
                //out.println("tEnd: " + tEnd);
                if (tStart.equals(tEnd))
                    break;
            }
            String tRep = t.substring(n - i);
            out.print(t);
            for (int j = 1; j < k; ++j) out.print(tRep);
            out.println();
        }

    }
}