D. Dreamoon and Sets
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 sisj 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.

Input

The single line of the input contains two space separated integers nk (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 .

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;
}
}



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