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：

 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 }

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