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.BufferedReader; 
import java.io.IOException; 
import java.io.InputStreamReader;

public class ManyEqualSubstrings{

    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 s = new FastReader();
        int t = 1;
        for(int u = 0;u<t;u++){
            int n = s.nextInt();
            int k = s.nextInt();
            String str = s.next();
            int i = 1;
            int j = n-1;
            int temp = -1;
            while(i<n){
                if(str.substring(i, n).equals(str.substring(0, j))){
                    temp = j-1;
                    break;
                }
                i++;
                j--;
            }
            StringBuilder sb = new StringBuilder();
            sb.append(str);
            for(i = 1;i<k;i++){
                sb.append(str.substring(temp+1, n));
            }
            System.out.println(sb.toString());
        }
    }

}