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

public class Many_Equal_Substrings {
//		http://codeforces.com/contest/1029/problem/A
	public static void main(String[] args) throws IOException{
		scanner sc=new scanner();
		int n=sc.nextInt(),k=sc.nextInt();
		String s=sc.next();
		StringBuilder sb=new StringBuilder();
		int l=0;
		for(int i=0;i<s.length()-1;i++) {
			if(s.substring(0, i+1).equals(s.substring(s.length()-i-1))) {
				sb=new StringBuilder(s.substring(s.length()-i-1));
			}
		}
		sb=new StringBuilder(s.substring(sb.length()));
		StringBuilder ss=new StringBuilder(s);
		for(int i=1;i<k;i++) {
			ss.append(sb.toString());
		}
		System.out.println(ss.toString());
	}
	
	static class scanner{
		BufferedReader br;
		StringTokenizer st;
		scanner(){
			br=new BufferedReader(new InputStreamReader(System.in));
		}
		String next() throws IOException {
			while(st==null||!st.hasMoreTokens())
				st=new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		int nextInt() throws NumberFormatException, IOException {
			return Integer.parseInt(next());
		}
	}
}