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:
//I AM THE CREED
/* //I AM THE CREED
/* package codechef; // don't place package name! */
import java.io.BufferedReader; 
import java.io.IOException; 
import java.io.InputStreamReader; 
import java.util.StringTokenizer; 
import java.util.*;
import java.awt.Point;
public class Main{
    
    public static void main(String[] args) throws IOException 
    { 
        
        FastScanner input = new FastScanner();
        int n=input.nextInt();
        int k=input.nextInt();
        String s=input.next();
        int max=n;
        for(int i=1;i<n;i++){
            if(s.substring(i).equals(s.substring(0, n-i))){
                max=i;
                break;
            }
        }
        StringBuilder sol=new StringBuilder();
        sol.append(s);
        k--;
        String part=s.substring(n-max);
        while(k!=0){
            sol.append(part);
            k--;
        }
        System.out.println(sol.toString());
    }
    static boolean isPrime(int n) 
    { 
        // Corner case 
        if (n <= 1) 
            return false; 
  
        // Check from 2 to n-1 
        for (int i = 2; i < n; i++) 
            if (n % i == 0) 
                return false; 
  
        return true; 
    } 
    static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
 
}