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.Scanner;

public class manyEqualSubstrings 
{
	int[] computePrefixSuffix(String s, int n)
	{
		int temp[] = new int[n];
		int index = 0;
		for(int i=1;i<n;)
		{
			if(s.charAt(i) == s.charAt(index))
			{
				temp[i] = index + 1;
				i++;
				index++;
			}
			else
			{
				if(index != 0)
					index = temp[index-1];
				else
				{
					temp[i]=0;
					i++;
				}
			}
		}
		return temp;			
	}
	public static void main(String[] args) 
	{
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int k = in.nextInt();
		in.nextLine();
		String str = in.nextLine();
		StringBuilder result = new StringBuilder();
		// Compute temporary array to maintain size of suffix which is same as prefix
		// Time/space complexity is O(size of pattern)
		
		manyEqualSubstrings ob = new manyEqualSubstrings();
		int temp[] = ob.computePrefixSuffix(str, n);
		for(int i=0;i<k;i++)
			result.append(str.substring(0, n-temp[n-1]));
		result.append(str.substring(n-temp[n-1]));
		System.out.println(result);
		in.close();
	}
}