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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

/**
 * @author Pavel Mavrin
 */
public class B {

    private void solve() throws IOException {
        int n = nextInt();
        int k = nextInt();
        boolean[] z = new boolean[100000];
        int[][] res = new int[n][4];
        for (int i = 0; i < n; i++) {
            res[i][0] = i * 6 + 1;
            res[i][1] = i * 6 + 3;
            res[i][2] = i * 6 + 5;
            int j = 2;
            while (z[j] || j % 3 == 0 || gcd(j, res[i][0]) > 1 ||  gcd(j, res[i][1]) > 1 ||  gcd(j, res[i][2]) > 1) {
                j += 2;
            }
            res[i][3] = j;
            for (int q = 0; q < 4; q++) {
                z[res[i][q]] = true;
            }
        }
        int m = 0;
        for (int i = 0; i < z.length; i++) {
            if (z[i]) m = i;
        }
        out.println(m * k);
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < 4; j++) {
                out.print(res[i][j] * k + " ");
            }
            out.println();
        }
    }

    private int gcd(int a, int b) {
        while (b > 0) {
            int c = a % b;
            a = b;
            b = c;
        }
        return a;
    }


    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st;
    PrintWriter out = new PrintWriter(System.out);

    String next() throws IOException {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(br.readLine());
        }
        return st.nextToken();
    }

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

    public static void main(String[] args) throws IOException {
        new B().run();
    }

    private void run() throws IOException {
        solve();
        out.close();
    }

}