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 Main {
	static int lps[];
	public static void main(String[] args) {
		FastReader sc = new FastReader();
			int n = sc.nextInt();
			int k = sc.nextInt();
			String s = sc.nextLine();
			lps = new int[n];
			computeLPSArray(s,n,lps);
			StringBuilder sb = new StringBuilder(s);
			for(int i=0;i<k-1;i++) {
				String sub = s.substring(lps[n-1], s.length());
				sb.append(sub);
			}
			System.out.println(sb.toString());
	}
	
	static void computeLPSArray(String pat, int M, int lps[])
    {
        // length of the previous longest prefix suffix
        int len = 0;
        int i = 1;
        lps[0] = 0; // lps[0] is always 0
 
        // the loop calculates lps[i] for i = 1 to M-1
        while (i < M)
        {
            if (pat.charAt(i) == pat.charAt(len))
            {
                len++;
                lps[i] = len;
                i++;
            }
            else // (pat[i] != pat[len])
            {
                // This is tricky. Consider the example.
                // AAACAAAA and i = 7. The idea is similar 
                // to search step.
                if (len != 0)
                {
                    len = lps[len-1];
 
                    // Also, note that we do not increment
                    // i here
                }
                else // if (len == 0)
                {
                    lps[i] = len;
                    i++;
                }
            }
        }
    }
}

class FastReader
{
    BufferedReader br;
    StringTokenizer st;

    public FastReader()
    {
        br = new BufferedReader(new
                 InputStreamReader(System.in));
    }

    String next()
    {
        while (st == null || !st.hasMoreElements())
        {
            try
            {
                st = new StringTokenizer(br.readLine());
            }
            catch (IOException  e)
            {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

    int nextInt()
    {
        return Integer.parseInt(next());
    }

    long nextLong()
    {
        return Long.parseLong(next());
    }

    double nextDouble()
    {
        return Double.parseDouble(next());
    }

    String nextLine()
    {
        String str = "";
        try
        {
            str = br.readLine();
        }
        catch (IOException e)
        {
            e.printStackTrace();
        }
        return str;
    }
}