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.*;
import java.util.*;

public class Q1 {

	public static void main(String args[]) throws IOException{
		FastReader sc = new FastReader();
		//int t = sc.nextInt();
		//for(int tc=0;tc<t;tc++) {
			
			int n = sc.nextInt();
			int m = sc.nextInt();
			String s = sc.next();
			
			int lps[] = new int[n];
			computeLPSArray(s , n , lps);
			
			int v = lps[n-1];
			StringBuilder sb = new StringBuilder("");
			if(v == 0) {
				for(int i=0;i<m;i++)
					sb.append(s);
				System.out.println(sb.toString());
			}
			else {
				sb.append(s);
				for(int i=0;i<m-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++;
                }
            }
        }
    }
 
	static 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;
		}
	}

}