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:

/*
 * @author romit17
 */
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;

public class A1029{
    
    void solve() throws IOException {
        PrintWriter out = new PrintWriter(System.out);
        StringBuilder sb = new StringBuilder("");        
        int n = ni(), k=ni();
        String s = ns();
        if(k==1)
        {
            System.out.println(s); return;
        }
        int m = 0;
        for(int i=1;i<n;i++)
        {
            String s1 = s.substring(0,i);
            String s2 = s.substring(n-i);
            if(s1.equals(s2))
                m = i;
        }
        
        if(m==0)
        {
            while(k-->0)
                sb.append(s);
            System.out.println(sb); return;
        }
        
        String ss = s.substring(m,n);
        sb.append(s);
        while(k-->1)
        {
            sb.append(ss);
        }
        System.out.println(sb);
    }
    
    public static void main(String[] args) throws IOException {
        new A1029().solve();
    }    
    
    private byte[] inbuf = new byte[1024];
    public int lenbuf = 0, ptrbuf = 0;    
    InputStream is = System.in;

    private int readByte() {
        if (lenbuf == -1) {
            throw new InputMismatchException();
        }
        if (ptrbuf >= lenbuf) {
            ptrbuf = 0;
            try {
                lenbuf = is.read(inbuf);
            } catch (IOException e) {
                throw new InputMismatchException();
            }
            if (lenbuf <= 0) {
                return -1;
            }
        }
        return inbuf[ptrbuf++];
    }
    
    private boolean isSpaceChar(int c) {
        return !(c >= 33 && c <= 126);
    }
    
    private int skip() {
        int b;
        while ((b = readByte()) != -1 && isSpaceChar(b));
        return b;
    }
    
    private double nd() {
        return Double.parseDouble(ns());
    }
    
    private char nc() {
        return (char) skip();
    }
    
    private String ns() {
        int b = skip();
        StringBuilder sb = new StringBuilder();
        while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != ' ')
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }
    
    private char[] ns(int n) {
        char[] buf = new char[n];
        int b = skip(), p = 0;
        while (p < n && !(isSpaceChar(b))) {
            buf[p++] = (char) b;
            b = readByte();
        }
        return n == p ? buf : Arrays.copyOf(buf, p);
    }
    
    private char[][] nm(int n, int m) {
        char[][] map = new char[n][];
        for (int i = 0; i < n; i++) {
            map[i] = ns(m);
        }
        return map;
    }
    
    private int[] na(int n) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = ni();
        }
        return a;
    }
    
    private int[] na1(int n) {
        int[] a = new int[n + 1];
        for (int i = 1; i < n + 1; i++) {
            a[i] = ni();
        }
        return a;
    }
    
    private long[] nb(int n) {
        long[] a = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = nl();
        }
        return a;
    }
    
    private long[] nb1(int n) {
        long[] a = new long[n + 1];
        for (int i = 1; i < n + 1; i++) {
            a[i] = nl();
        }
        return a;
    }
    
    private int ni() {
        int num = 0, b;
        boolean minus = false;
        while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        
        while (true) {
            if (b >= '0' && b <= '9') {
                num = num * 10 + (b - '0');
            } else {
                return minus ? -num : num;
            }
            b = readByte();
        }
    }
    
    private long nl() {
        long num = 0;
        int b;
        boolean minus = false;
        while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        
        while (true) {
            if (b >= '0' && b <= '9') {
                num = num * 10 + (b - '0');
            } else {
                return minus ? -num : num;
            }
            b = readByte();
        }
    }
    
}