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.math.*;
import java.util.*;
import java.io.*;

public class _Main {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;
        public FastReader() {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }
        String next() {
            while (st == null || !st.hasMoreElements()){
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException  e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt() {
            return Integer.parseInt(next());
        }
        long nextLong() {
            return Long.parseLong(next());
        }
        double nextDouble() {
            return Double.parseDouble(next());
        }
        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
    static boolean[] sieve(int N){
        boolean[] prime = new boolean[N + 1];
        java.util.Arrays.fill(prime, true);
        prime[0] = false;
        prime[1] = false;
        for (int i = 2; i*i < N + 1; i++){
            if (prime[i]) {
                for (int j = 2; i*j < N+1; j++) {
                    prime[i*j] = false;
                }
            }
        }
        return prime;
    }
    public static int product(int x){
        String temp = String.valueOf(x);
        char[] Digits = temp.toCharArray();
        int length = Digits.length;
        int ans = 1;
        for (int i = 0; i < length; i++){
            ans *= Character.getNumericValue(Digits[i]);
        }
        return ans;
    }
    public static int FindSmallest (int [] arr){
        int index = 0;
        int min = arr[index];
        for (int i=1; i<arr.length; i++){
            if (arr[i] < min ){
                min = arr[i];
                index = i;
            }
        }
        return index ;

    }
    public static void main(String[] args) {
        FastReader S = new FastReader();
        int n = S.nextInt();
        int k = S.nextInt();
        String str = S.next();
        String ans = new String();
        ans += str;
        for (int i = 1; i < k; i++){
            int index = n - 1;
            for (int j = 0; j < n - 1; j++){
                if (ans.substring(0, n - j - 1).equals(str.substring(j + 1, n))) {
                    break;
                } else index--;
            }
            //System.out.println("index = " + index);
            if (index == 0){
                ans += str;
            } else {
                for (int j = index; j < n; j++) {
                    ans += str.charAt(j);
                }
            }

//            System.out.println("i = " + i);
            //System.out.println(ans);
        }
        System.out.println(ans);
    }
}