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.*;

public class codechef
{
	public static void main(String[] args) throws IOException
	{
		FastReader in=new FastReader();
		int n=in.nextInt();
		int k=in.nextInt();
		char[] s=in.next().toCharArray();
		int p=0;
		for(int i=1;i<s.length;i++)
		{
			int j=i,a=0;
			p=0;
			while(j<s.length)
			{
				if(s[a]!=s[j])
					break;
				j++;
				a++;
				p++;
			}
			if(j==s.length)
				break;
		}
		for(int i=0;i<n;i++)
			System.out.print(s[i]);
		for(int i=1;i<k;i++)
		{
			for(int j=p;j<n;j++)
				System.out.print(s[j]);
		}
	}
}

class FastReader
{
	BufferedReader br;
	StringTokenizer st;
 
	public FastReader()
	{
		br=new BufferedReader(new InputStreamReader(System.in));
	}
 
	public String next() throws IOException
	{
		if(st==null || !st.hasMoreElements())
		{
			st=new StringTokenizer(br.readLine());
		}
		return st.nextToken();
	}
 
	public int nextInt() throws IOException
	{
		return Integer.parseInt(next());
	}
 
	public long nextLong() throws IOException
	{
		return Long.parseLong(next());
	}
}