Codeforces Round #272 (Div. 2) D. Dreamoon and Sets 构造

D. Dreamoon and Sets

题目连接：

http://www.codeforces.com/contest/476/problem/D

Description

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 from S, .

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 Input

1 1

Sample Output

5

1 2 3 5

Hint

题意

让你构造n个集合，使得每个集合里面都有四个元素，且两两元素之间的gcd都是k

让你使得其中最大数最小

题解：

其实就是样例的构造方法，这个得自己手玩一玩才行

代码

#include<bits/stdc++.h>
using namespace std;

int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    printf("%d\n",(n-1)*k*6+5*k);
    for(int i=0;i<n;i++)
        printf("%d %d %d %d\n",i*k*6+k,i*k*6+2*k,i*k*6+3*k,i*k*6+5*k);
}