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.*;
import java.util.StringTokenizer;

public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskA solver = new TaskA();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskA {
        int t, k;

        public void solve(int testNum, InputReader in, PrintWriter out) {
//            while(testNum-- >= 0) {
                t = in.nextInt();
                k = in.nextInt();
                String substr = in.next();

                int flag = -1;

                for (int i = 1; i < t; ++i) {
                    if (substr.substring(i, t).equals(substr.substring(0, t - i))) {
                        flag = i;
                        break;
                    }
                }

                String remain = null;

                if (flag != -1) {
                    remain = substr.substring(t - flag, t);
                } else {
                    remain = substr;
                }

                String res = substr;

                for (int i = 0; i < k-1; ++i) {
                    res += remain;
                }

                out.println(res);
            }
//        }
    }

    static class InputReader {
        private BufferedReader reader;
        private StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (Exception e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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