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 javax.management.DynamicMBean;
import javax.print.DocFlavor;
import javax.print.attribute.HashAttributeSet;
import java.awt.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.nio.file.attribute.FileAttribute;
import java.util.List;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.*;
import java.util.Arrays;
import java.util.concurrent.Callable;



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;
        }
    }

    public static void main(String[] args) {
        FastReader sc = new FastReader();
       int n=sc.nextInt();
       int k=sc.nextInt();
       String s=sc.next();
       int pos=0;
       for(int i=n-1;i>=1;i--){
           String str=s.substring(0,i);
           if(s.endsWith(str)){
               pos=i;
               break;
           }
       }
       String str=s.substring(pos,n);
       for(int j=0;j<k-1;j++){
           s+=str;
       }
       System.out.println(s);
    }
}