#include <bits/stdc++.h>
using namespace std;
int n, m;
int main() {
  scanf("%d%d", &n, &m);
  printf("%d\n", m * (6 * n - 1));
  for (int i = 1; i <= n; i++) {
    int t = (i - 1) * 6 + 1;
    printf("%d %d %d %d\n", t * m, (t + 1) * m, (t + 2) * m, (t + 4) * m);
  }
  return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

Dreamoon likes to play with sets, integers and <image>. <image> 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, <image>.

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 <image>.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m;
int main() {
  scanf("%d%d", &n, &m);
  printf("%d\n", m * (6 * n - 1));
  for (int i = 1; i <= n; i++) {
    int t = (i - 1) * 6 + 1;
    printf("%d %d %d %d\n", t * m, (t + 1) * m, (t + 2) * m, (t + 4) * m);
  }
  return 0;
}

```