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.
Example
Input
1 1
Output
5
1 2 3 5
Input2 2Output222 4 6 2214 18 10 16Note
For the first example it's easy to see that set {1, 2, 3, 4} isn't a valid set of rank 1 since .
一题找规律的题，开始也不会做，同学教我的，
每一行找互质的最小数字就得了
1 2 3 5
7 8 9 11
13 14 15 17
。。。。。。
。。。。。。
x x +1 x+2 x+4(x要是奇数才行）
代码：
#include<iostream>#include<algorithm>#include<stdio.h>#include<string.h>#include<stdlib.h>#include<set>using namespace std;int main(){ int n,m; while(~scanf("%d%d",&n,&m)) { printf("%d\n",(6*n-1)*m); for(int i=1;i<=m;i++){ printf("%d ",(6*i-5)*m); printf("%d ",(6*i-4)*m); printf("%d ",(6*i-3)*m); printf("%d\n",(6*i-1)*m); } } return 0;}