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.*;
import java.text.*;
import java.math.*;
public class Main{
	public static StringTokenizer token = new StringTokenizer("");
	public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	public static PrintWriter output;
	public static void main(String[] args) throws Exception {
		int N = ni();
		int K = ni();
		String str = ns();
		String temp = "";
		int a = N;
		for(int i=N-1;i>0;i--){
			temp = str.substring(i,N);
			int x = str.indexOf(temp);
			if(x==0){
				a = i;
			}
			if(x!=-1 && x<i){

			}
			else{
				break;
			}
		}
		for(int i=0;i<K;i++){
			System.out.print(str.substring(0,a));	
		}
		System.out.println(str.substring(a,N));
		
	}




	//Scanning Inputs and Converting data library
	//Parse Integer
	public static Integer pi(String str){
		return Integer.parseInt(str);
	}
	//Parse Long
	public static Long pl(String str){
		return Long.parseLong(str);
	}
	//Taking Integer Input
	public static Integer ni() throws IOException{
		if(!token.hasMoreElements()){
			token = new StringTokenizer(br.readLine());
		}
		return Integer.parseInt(token.nextToken());
	}
	//Taking Long Input
	public static Long nl() throws IOException{
		if(!token.hasMoreElements()){
			token = new StringTokenizer(br.readLine());
		}
		return Long.parseLong(token.nextToken());
	}
	//Taking String Input
	public static String ns() throws IOException{
		return new StringTokenizer(br.readLine()).nextToken();
	}
	//Taking Integer Array
	public static int[] ai(int A[]) throws IOException{
		token = new StringTokenizer(br.readLine());
		for(int i=0;i<A.length;i++){
			A[i] = pi(token.nextToken());
		}
		return A;
	}
	//Taking String Array
	public static String[] as(String S[]) throws IOException{
		token = new StringTokenizer(br.readLine());
		for(int i=0;i<S.length;i++){
			S[i] = token.nextToken();
		}
		return S;
	}
}