#include <bits/stdc++.h>
using namespace std;
long n, m;
long a, b;
long gcd(int a, int b) {
  int c;
  while (a != 0) {
    c = a;
    a = b % a;
    b = c;
  }
  return b;
}
int main() {
  cin >> m >> n;
  cout << (5 + 6 * m - 6) * n << endl;
  cout << 1 * n << ' ' << 2 * n << ' ' << 3 * n << ' ' << 5 * n << endl;
  long pizd = 4, t = 7;
  for (int k = 1; k <= m - 1; ++k) {
    while (gcd(pizd, t) != 1 || gcd(pizd, t + 2) != 1 || gcd(pizd, t + 4) != 1)
      pizd += 2;
    cout << n * pizd << ' ' << n * t << ' ' << n * (t + 2) << ' ' << n * (t + 4)
         << endl;
    pizd += 2;
    t += 6;
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
long n, m;
long a, b;
long gcd(int a, int b) {
  int c;
  while (a != 0) {
    c = a;
    a = b % a;
    b = c;
  }
  return b;
}
int main() {
  cin >> m >> n;
  cout << (5 + 6 * m - 6) * n << endl;
  cout << 1 * n << ' ' << 2 * n << ' ' << 3 * n << ' ' << 5 * n << endl;
  long pizd = 4, t = 7;
  for (int k = 1; k <= m - 1; ++k) {
    while (gcd(pizd, t) != 1 || gcd(pizd, t + 2) != 1 || gcd(pizd, t + 4) != 1)
      pizd += 2;
    cout << n * pizd << ' ' << n * t << ' ' << n * (t + 2) << ' ' << n * (t + 4)
         << endl;
    pizd += 2;
    t += 6;
  }
  return 0;
}

```