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.util.*;
import java.io.*;
import java.text.*;

public class Main {
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in), 32768);
    static PrintWriter out = new PrintWriter(System.out);
    static StringTokenizer t;

    static String sn() {
        while (t == null || !t.hasMoreTokens()) {
            try {
                t = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                throw new RuntimeException();
            }
        }
        return t.nextToken();
    }

    static int ni() {
        return Integer.parseInt(sn());
    }

    static long nlo() {
        return Long.parseLong(sn());
    }

    public static void main(String args[]) {
        int t = 1;
        while (t-- > 0) {
            solve();
        }
        out.close();
    }

    static long mod = 998244353;

    static long pro(long a, long b) {
        return (a % mod * b % mod) % mod;
    }

    static long add(long a, long b) {
        return (a + b) % mod;
    }

    static int maxn = 200005;

    static int pow(int n) {
        return (int) (Math.log10(n) / Math.log10(2));
    }

    static void solve() {
        int n = ni();
        int k = ni();
        String t = "";
        char a[] = (t = sn()).toCharArray();
        int kmp[] = new int[n];
        int i = 1, j = 0;
        while (i < n) {
            if (a[i] == a[j]) {
                kmp[i] = j + 1;
                i++;
                j++;
            } else {
                if (j != 0)
                    j = kmp[j - 1];
                else {
                    kmp[i] = 0;
                    i++;
                }
            }
        }
//        out.println(Arrays.toString(kmp));
        String s = t.substring(kmp[n - 1]);
        StringBuilder sb = new StringBuilder(t);
        k -= 1;
        while (k-- > 0)
            sb.append(s);
        out.println(sb);
    }
}