#include <bits/stdc++.h>
const int inf = 1 << 30;
const long long INF = 1ll << 61;
const double pi = acos(-1.0);
using namespace std;
int n, k;
void init() {}
bool input() {
  while (cin >> n >> k) {
    return false;
  }
  return true;
}
void cal() {
  int maxn = (6 * (n - 1) + 5) * k;
  cout << maxn << endl;
  for (int i = 0; i < n; i++) {
    printf("%d %d %d %d\n", 6 * i * k + k, 6 * i * k + 2 * k, 6 * i * k + 3 * k,
           6 * i * k + 3 * k + 2 * k);
  }
}
void output() {}
int main() {
  while (true) {
    init();
    if (input()) return 0;
    cal();
    output();
  }
  return 0;
}

### Prompt

Generate a CPP solution to 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>
const int inf = 1 << 30;
const long long INF = 1ll << 61;
const double pi = acos(-1.0);
using namespace std;
int n, k;
void init() {}
bool input() {
  while (cin >> n >> k) {
    return false;
  }
  return true;
}
void cal() {
  int maxn = (6 * (n - 1) + 5) * k;
  cout << maxn << endl;
  for (int i = 0; i < n; i++) {
    printf("%d %d %d %d\n", 6 * i * k + k, 6 * i * k + 2 * k, 6 * i * k + 3 * k,
           6 * i * k + 3 * k + 2 * k);
  }
}
void output() {}
int main() {
  while (true) {
    init();
    if (input()) return 0;
    cal();
    output();
  }
  return 0;
}

```