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 static java.lang.Double.parseDouble;
import static java.lang.Integer.MAX_VALUE;
import static java.lang.Integer.max;
import static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.Math.min;
import static java.lang.System.exit;

import java.io.*;
import java.util.*;
import java.util.concurrent.atomic.AtomicBoolean;

public class A {

    static BufferedReader in;
    static PrintWriter out;
    static StringTokenizer tok;

    static void solve() throws Exception {
        int n = nextInt(), k = nextInt();
        String s = next();
        String left = s, right = "";

        for(int i = 1; i < n; i++) {
            String x = s.substring(i,n);
            if(s.startsWith(x)) {
                left = s.substring(0,i);
                right = x;
                break;
            }
        }
        for(int i = 0; i < k; i++) out.print(left);
        if(!s.equals(left)) out.print(right);
    }

    static boolean exist(int a[], int x) {
        return Arrays.binarySearch(a, x) >= 0;
    }

    private static long gcd(long a, long b) {
        while (b > 0) {
            long temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }

    private static long lcm(long a, long b) {
        return a * (b / gcd(a, b));
    }

    static int nextInt() throws IOException {
        return parseInt(next());
    }

    static long nextLong() throws IOException {
        return parseLong(next());
    }

    static double nextDouble() throws IOException {
        return parseDouble(next());
    }

    static String next() throws IOException {
        while (tok == null || !tok.hasMoreTokens()) {
            tok = new StringTokenizer(in.readLine());
        }
        return tok.nextToken();
    }

    public static void main(String[] args) throws Exception {
        try {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(new OutputStreamWriter(System.out));
            //long lStartTime = System.currentTimeMillis();
            solve();
            //long lEndTime = System.currentTimeMillis();
            //out.println("Elapsed time in seconds: " + (double)(lEndTime - lStartTime) / 1000.0);
            in.close();
            out.close();
        } catch (Throwable e) {
            e.printStackTrace();
            exit(1);
        }
    }
}