Codeforces Round #272 (Div. 1)B（構造）

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, sjfrom 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

題意：構造出n個集合，每個集合由四個數字組成，集合的數滿足兩兩的GCD等於K，且要求集合裏的數的最大值最小

思路：構造

            1 2 3 5

            7 8 9 11

            13 14 15 17

            ........

            1+6*k  2+6*k  3+6*k  5+6*k

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

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