CodeForces

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 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.

Examples

input

1 1

output

51 2 3 5

input

2 2

output

222 4 6 2214 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 .

题意：输入 n,k 然后让你在 1 到 m （题目要求m尽可能的小）中找出n个集合，每个集合有四个数 且这四个数的gcd为k（所有数字不重复）。最终输出m，和n个集合（集合的顺序和集合中的整数的顺序都不重要） 。

思路：首先想到公式  若gcd(a,b)==1则，gcd(a*k,b*k)==k; 然后此时可以直接去寻找 k==1的情况。可以找到规律 1,2,3,5  7,8,9,11  13,14,15,17  .... 

代码：

 

#include <cstdio>#include <iostream>#define ll long longusing namespace std;ll ans[10005][4];void init(){    ll j=1;    for(int i=0;i<=10000;i++)    {        ans[i][0]=j++;        ans[i][1]=j++;        ans[i][2]=j++;        j++;        ans[i][3]=j++;        j++;    }}int main(){    int n,k;    init();    while(~scanf("%d %d",&n,&k))    {        printf("%lld\n",ans[n-1][3]*k);        for(int i=0;i<n;i++)        {            printf("%lld %lld %lld %lld\n",ans[i][0]*k,ans[i][1]*k,ans[i][2]*k,ans[i][3]*k);        }    }    return 0;}