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 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 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：

 1 #include<cstdio>
 2 #include<cstring>
 3 using namespace std;
 4 const int maxn=10005;
 5 __int64 ans[maxn][4];
 6 void work()
 7 {
 8    __int64 k=1;
 9   for(int i=0;i<10000;i++)
10   {
11       ans[i][0]=k++;
12       ans[i][1]=k++;
13       ans[i][2]=k++;
14       ans[i][3]=++k;
15       k+=2;
16   }
17 }
18 int main()
19 {
20   int n,k;
21   work();
22   while(scanf("%d%d",&n,&k)!=EOF)
23   {
24       printf("%I64d\n",ans[n-1][3]*k);
25       for(int i=0;i<n;i++)
26         printf("%I64d %I64d %I64d %I64d\n",ans[i][0]*k,ans[i][1]*k,ans[i][2]*k,ans[i][3]*k);
27   }
28  return 0;
29 }