Codeforces Round #272 (Div. 2) D Dreamoon and Sets（构造）

思路：需要手玩一会....

#include<bits/stdc++.h>
using namespace std;

int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    printf("%d\n",(n-1)*k*6+5*k);
    for(int i=0;i<n;i++)
        printf("%d %d %d %d\n",i*k*6+k,i*k*6+2*k,i*k*6+3*k,i*k*6+5*k);
}

D. 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.

Examples

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 .