Dreamoon likes to play with sets, integers and .  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 fromS, .

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.

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

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.

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

构造。当k等于1时，推几组数据。比如1,2,3,5；7,8,9,11；13,14,15,17。19,20,21,23；25,26,27,29。就会发现是以6为周期，而对每一个周期内的数乘以k就会使周期内的数两两的最大公约数为k。

代码：

#include <cstdio>

#include <iostream>

#include <cstring>

using namespace std;

int main()

{

    int n, k;

    scanf("%d %d", &n, &k);

    int a = 1, b = 2, c = 3, d = 5;

    printf("%d\n", (d * k + 6 * k * (n- 1)));

    a*=k;

    b*=k;

    c*=k;

    d*=k;

    for(int i = 0; i < n; i++)

    {

        printf("%d %d %d %d\n",a, b, c, d);

        a += 6 * k;

        b += 6 * k;

        c += 6 * k;

        d += 6 * k;

    }

}