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.*;
 
public class Main {
    public static void main(String[] args) throws Exception {
        FastScanner sc = new FastScanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        StringBuilder sb = new StringBuilder();
        int len = sc.nextInt();
        int rep = sc.nextInt();
        char[] s = sc.next().toCharArray();
        int pref = -1;
        for(int i = 1; i < len; i++){
            boolean flg = true;
            for(int j = 0; j < len-i; j++){
                if(s[i+j] != s[j]){
                    flg = false;
                    break;
                }
            }
            if(flg){
                pref = len-i;
                break;
            }
        }
        sb.append(s);
        for(int i = 1; i < rep; i++){
            if(pref != -1){
                for(int j = pref; j < len; j++){
                    sb.append(s[j]);
                }
            }else{
                sb.append(s);
            }
        }
        pw.println(sb.toString());
        sb.setLength(0);
        pw.flush();
    }

    static class GeekInteger {
        public static void save_sort(int[] array) {
            shuffle(array);
            Arrays.sort(array);
        }
 
        public static int[] shuffle(int[] array) {
            int n = array.length;
            Random random = new Random();
            for (int i = 0, j; i < n; i++) {
                j = i + random.nextInt(n - i);
                int randomElement = array[j];
                array[j] = array[i];
                array[i] = randomElement;
            }
            return array;
        }
 
    }
}

class FastScanner {
    private BufferedReader reader = null;
    private StringTokenizer tokenizer = null;
    public FastScanner(InputStream in) {
        reader = new BufferedReader(new InputStreamReader(in));
        tokenizer = null;
    }

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

    public String nextLine() {
        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                return reader.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken("\n");
    }

    public long nextLong() {
        return Long.parseLong(next());
    }

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

    public double nextDouble() {
         return Double.parseDouble(next());
    }
    
    public String[] nextArray(int n) {
        String[] a = new String[n];
        for (int i = 0; i < n; i++)
            a[i] = next();
        return a;
    }

    public int[] nextIntArray(int n) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++)
            a[i] = nextInt();
        return a;
    }

    public long[] nextLongArray(int n) {
        long[] a = new long[n];
        for (int i = 0; i < n; i++)
            a[i] = nextLong();
        return a;
    } 
}