cf(#div1 B. Dreamoon and Sets)(数论)

B. 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 s_i, s_j 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 test(s)

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 .

题意： 给你任意n,k,要你求出n组gcd(si,sj)=k的四个元素的组合.........

其实对于gcd(a,b)=k,我们只需要求出gcd(a,b)=1;然后进行gcd(a,b)*k=k;

不难发现这些数是固定不变的而且还有规律可循，即1,2,3,5    7 8 9 11   13 14 15 17   每一个段直接隔着2,段内前3个连续，后一个隔着2.....

代码：

#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=10005;
__int64 ans[maxn][4];
void work()
{
   __int64 k=1;
  for(int i=0;i<10000;i++)
  {
      ans[i][0]=k++;
      ans[i][1]=k++;
      ans[i][2]=k++;
      ans[i][3]=++k;
      k+=2;
  }
}
int main()
{
  int n,k;
  work();
  while(scanf("%d%d",&n,&k)!=EOF)
  {
      printf("%I64d
",ans[n-1][3]*k);
      for(int i=0;i<n;i++)
        printf("%I64d %I64d %I64d %I64d
",ans[i][0]*k,ans[i][1]*k,ans[i][2]*k,ans[i][3]*k);
  }
 return 0;
}