Dreamoon and Sets CodeForces - 476D

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Dreamoon and Sets

Dreamoon likes to play with sets, integers and $\text{[math]}$ . $\text{[math]}$ 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 s_i, s_j from S, $\text{[math]}$ .

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

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 $\text{[math]}$ .

题意：

S为含四个整数元素的集合，满足S中任意2个元素si，sj 满足gcd（si,sj）=k；求n个S集合（所有数字只能用一次）先输出n个S集合中最小的最大值再输出n个任意满足条件的S集合。

思路：

找到gcd（a,b）=1的S集合，因为gcd(a,b)=1→gcd(a*k,b*k)=k; 所以把S集合同乘k即可。

可见S集合满足S中的最大值为最小的 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 可得出规律

1 2 3 4 5 6

7 8 9 10 11 12

13 14 15 16 17 18

19 20 21 22 2324

AC：

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
typedef long long LL;
using namespace std;

int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    int maxn=(1+(n-1)*6+4)*k;
    printf("%d\n",maxn);
    int dd=0;
    for(int i=1;i<maxn&&dd<n;i+=6)
    {
        int a,b,c,d;
        a=i*k;
        b=(i+1)*k;
        c=(i+2)*k;
        d=(i+4)*k;
        printf("%d %d %d %d\n",a,b,c,d);
        dd++;
    }
}