D. Dreamoon and Sets（Codeforces Round #272)

Dreamoon likes to play with sets, integers and $\text{[math]}$ . $\text{[math]}$ 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, $\text{[math]}$ .

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.

Sample test(s)

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 $\text{[math]}$ .

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
", (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
",a, b, c, d);
        a += 6 * k;
        b += 6 * k;
        c += 6 * k;
        d += 6 * k;
    }
}