#include <bits/stdc++.h>
using namespace std;
const long long maxn = 1e6 + 10;
const long long mod = 1e9 + 7;
long long h[maxn];
long long dp[maxn];
signed main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  long long n, k;
  cin >> n >> k;
  long long plus[] = {1, 2, 3, 5};
  cout << (6 * (n - 1) + 5) * k << "\n";
  for (long long i = 0; i < n; ++i) {
    for (long long j = 0; j < 4; ++j) {
      cout << (6 * i + plus[j]) * k << " ";
    }
    cout << "\n";
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
const long long maxn = 1e6 + 10;
const long long mod = 1e9 + 7;
long long h[maxn];
long long dp[maxn];
signed main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  long long n, k;
  cin >> n >> k;
  long long plus[] = {1, 2, 3, 5};
  cout << (6 * (n - 1) + 5) * k << "\n";
  for (long long i = 0; i < n; ++i) {
    for (long long j = 0; j < 4; ++j) {
      cout << (6 * i + plus[j]) * k << " ";
    }
    cout << "\n";
  }
}

```