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;
        FastScanner in = new FastScanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        B solver = new B();
        solver.solve(1, in, out);
        out.close();
    }

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

            String s = in.nextLine();

            int st = 1;
            int id = -1;
            while (true) {
                for (; st < n; st++) {
                    if (s.charAt(st) == s.charAt(0)) break;
                }
                if (st == n) break;
                int x = st;
                for (; x < n; x++) {
                    if (s.charAt(x) != s.charAt(x - st)) break;
                }
                if (x == n) {
                    id = st;
                    break;
                }
                st = st + 1;
            }
            if (id == -1) {
                for (int i = 0; i < k; i++) {
                    out.print(s);
                }
                return;
            }

            String sub = s.substring(n - id);
            out.print(s);
            for (int i = 1; i < k; i++) {
                out.print(sub);
            }
        }

    }

    static class FastScanner {
        private BufferedReader bufferedReader;
        private StringTokenizer stringTokenizer;

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

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

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

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

    }
}