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>.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long int scan() {
  long long int t = 0;
  char c, ch;
  ch = getchar();
  if (ch == '-') {
    c = getchar();
  } else {
    c = ch;
  }
  while (c < '0' || c > '9') {
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    t = (t << 3) + (t << 1) + c - '0';
    c = getchar();
  }
  if (ch == '-') {
    return -t;
  } else {
    return t;
  }
}
int main() {
  ios::sync_with_stdio(false);
  long long int i, j, k, t, n, m, x, y;
  cin >> n >> m;
  cout << (5 + 6 * (n - 1)) * m << "\n";
  for (i = 0; i < n; i++) {
    cout << (1 + 6 * (i)) * m << " " << (2 + 6 * (i)) * m << " "
         << (3 + 6 * (i)) * m << " " << (5 + 6 * (i)) * m << "\n";
  }
  return 0;
}