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
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        ReaderFastIO in = new ReaderFastIO(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, ReaderFastIO in, PrintWriter out) {

            int n = in.nextInt();
            int k = in.nextInt();
            String t = in.nextLine();
            String r = new String(t);

            for (int i = 1; i < t.length(); i++) {
                if (t.startsWith(t.substring(i, t.length()))) {
                    r = t.substring(t.length() - i, t.length());
                    break;
                }
            }

            out.print(t);

            for (int i = 1; i < k; i++) {
                out.print(r);
            }

            out.println();

        }

    }

    static class ReaderFastIO {
        BufferedReader br;
        StringTokenizer st;

        public ReaderFastIO() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public ReaderFastIO(InputStream input) {
            br = new BufferedReader(new InputStreamReader(input));
        }

        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());
        }

        public String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

    }
}