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Codeforces Round #272 (Div. 2) D. Dreamoon and Sets 构造
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发布时间:2019-06-28

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D. Dreamoon and Sets

题目连接:

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
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

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