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.util.Scanner;
import java.io.OutputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author himalayjoriwal@gmail.com
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskD solver = new TaskD();
        solver.solve(1, in, out);
        out.close();
    }
}

class TaskD {
    public void solve(int testNumber, Scanner in, PrintWriter out) {
        int n = in.nextInt();
        int k = in.nextInt();

        long ans[][] = new long[n][4];
        ans[0][0] = k; ans[0][1] = 2*k; ans[0][2] = 3*k; ans[0][3] = 5*k;
        long m = 5*k;
        long start = 11;
       long tmp[] = new long[4];

        for (int i = 1; i < n; i++) {

            boolean got = false;
            while (!got) {
                getlast4(tmp, start);
                if (tmp[0]*k > m) {
                    got = true;
                    for (int j = 0; j < 4; j++) {
                        ans[i][j] = tmp[j] * k;
                    }
                    start += 4;
                    m = ans[i][3];
                }else
                    start++;
            }
        }
        out.println(m);
        StringBuffer sb = new StringBuffer();
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < 3; j++)
                sb.append(ans[i][j] + " ");
            sb.append(ans[i][3] + "\n");
        }
        out.println(sb.toString());
    }

    private void getlast4(long[] tmp, long start) {

        tmp[3] = start;

        for (int i = 2; i > -1; i--) {

            boolean got = true;
            do {
                tmp[i] = --start;
                got = true;
                for (int j = i + 1; j < 4; j++) {
                    if (gcd(tmp[j], tmp[i]) != 1)
                        got = false;
                }

            }while (!got);

        }

    }

    long gcd(long a, long b)
    {
        if (a == 0 || b == 0)
            return a+b;
        return gcd(b%a, a);
    }
}