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

public class ManyEqualSubstrings {
	public static BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
	public static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
	public static StringTokenizer st;
	
	public static void main(String[] args) throws IOException {
		int n = nextInt();
		int k = nextInt();
		String s = nextString();
		
		int m = n;
		for (int i = n-1; i > 0; i--)
			if (s.substring(i, n).equals(s.substring(0, n-i)))
				m = i;
		
		for (int i = 0; i < k; i++)
			System.out.print(s.substring(0, m));
		System.out.println(s.substring(m));
	}
	
	public static String nextLine() throws IOException {
		return in.readLine();
	}

	public static String nextString() throws IOException {
		while (st == null || !st.hasMoreTokens())
			st = new StringTokenizer(in.readLine());
		return st.nextToken();
	}

	public static int nextInt() throws IOException {
		return Integer.parseInt(nextString());
	}

	public static long nextLong() throws IOException {
		return Long.parseLong(nextString());
	}

	public static int[] intArray(int n) throws IOException {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = nextInt();
		return a;
	}

	public static int[][] intArray(int n, int m) throws IOException {
		int[][] a = new int[n][m];
		for (int i = 0; i < n; i++)
			for (int j = 0; j < m; j++)
				a[i][j] = nextInt();
		return a;
	}

	public static long[] longArray(int n) throws IOException {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nextLong();
		return a;
	}
}