Problem: Dreamoon likes to play with sets, integers and <image>. <image> is defined as the largest positive integer that divides both a and b.

Let S be a set of exactly four distinct integers greater than 0. Define S to be of rank k if and only if for all pairs of distinct elements si, sj from S, <image>.

Given k and n, Dreamoon wants to make up n sets of rank k using integers from 1 to m such that no integer is used in two different sets (of course you can leave some integers without use). Calculate the minimum m that makes it possible and print one possible solution.

Input

The single line of the input contains two space separated integers n, k (1 ≤ n ≤ 10 000, 1 ≤ k ≤ 100).

Output

On the first line print a single integer — the minimal possible m. 

On each of the next n lines print four space separated integers representing the i-th set.

Neither the order of the sets nor the order of integers within a set is important. If there are multiple possible solutions with minimal m, print any one of them.

Examples

Input

1 1


Output

5
1 2 3 5


Input

2 2


Output

22
2 4 6 22
14 18 10 16

Note

For the first example it's easy to see that set {1, 2, 3, 4} isn't a valid set of rank 1 since <image>.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.math.BigInteger;

@SuppressWarnings("unused")
public class Main {
    public static void main(String[] args) throws FileNotFoundException {
        InputStream inputStream = System.in;
        // InputStream inputStream = new FileInputStream(new File(
        // "C:\\Users\\Administrator\\Codeblocks\\First\\1.in"));
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskA solver = new TaskA();
        solver.solve(1, in, out);
        out.close();
    }
}

// 使用对象时,首先考虑是否要采用深复制
class TaskA {
    static final int mod = (int) 1e9 + 7;
    static final int INF = 0x7fffffff;
    static final double eps = 1e-7;
    static final int maxn = (int) 1e6;

    // 声明全局变量

    public void solve(int testNumber, InputReader in, PrintWriter out) {
        int N = in.nextInt();
        int K = in.nextInt();
        int a = 1, b = 2, c = 3, d = 5;
        out.println((d + 6 * (N - 1)) * K);
        a *= K;
        b *= K;
        c *= K;
        d *= K;
        for (int i = 0; i < N; i++) {
            out.print(a + " ");
            a += K * 6;
            out.print(b + " ");
            b += K * 6;
            out.print(c + " ");
            c += K * 6;
            out.print(d + "\n");
            d += K * 6;
        }
    }

}

class InputReader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;

    public InputReader(InputStream stream) {
        reader = new BufferedReader(new InputStreamReader(stream), 32768);
        tokenizer = null;
    }

    public String next() {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                tokenizer = new StringTokenizer(reader.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken();
    }

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

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

}