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.*;
public class subString {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		int n=sc.nextInt();
		int t=sc.nextInt();
		String s=sc.next();
		int x=lpsu(s);
		String a=s.substring(x, n);
		if(x==n-1){
			for(int i=0;i<t-1;i++){
				s+=s.charAt(0);
			}
			//System.out.println(s);
		}
		else
		for(int i=0;i<t-1;i++){
			s+=a;
		}
		System.out.println(s);

	}
	public static int lpsu(String s){
		int[] lps=new int[s.length()];
		int i=1;
		int len=0;
		lps[len]=0;
		while(i<s.length()){
			if(s.charAt(i)==s.charAt(len)){
				len++;
				lps[i]=len;
				i++;
			}
			else{
				if(len!=0){
					len=lps[len-1];
				}
				else{
					lps[i]=0;
					i++;
				}
			}
		}
		/*
		if(lps[s.length()-1]==s.length()){
			return lps[s.length()-1];
		}*/
		//System.out.println(lps[s.length()-1]);
		return lps[s.length()-1];
	}

}