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.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author bacali
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastScanner in = new FastScanner(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, FastScanner in, PrintWriter out) {
            int n = in.nextInt();
            int m = in.nextInt();
            String ch = in.next();
            for (int i = 0; i < n; i++) {
                String ch2 = ch.substring(0, i + 1);
                StringBuilder b = new StringBuilder();
                for (int j = 0; j < m - 1; j++) {
                    b.append(ch2);
                }
                b.append(ch);
                String res = b.toString();
                int c = 0;
                int ind = 0;
                while ((ind = res.indexOf(ch, ind)) != -1) {
                    c++;
                    ind++;
                }
                if (c == m) {
                    out.println(res);
                    return;
                }
            }
        }

    }

    static class FastScanner {
        private BufferedReader br;
        private StringTokenizer st;

        public FastScanner(InputStream inputStream) {
            br = new BufferedReader(new InputStreamReader(inputStream));
        }

        public String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

    }
}