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;
const int d[] = {1, 2, 3, 5};
void solve() {
  int n, k;
  scanf("%d%d", &n, &k);
  printf("%d\n", (6 * (n - 1) + 5) * k);
  for (int i(0), _l((int)(n)-1); i <= _l; ++i) {
    for (int j(0), _l((int)(4) - 1); j <= _l; ++j)
      printf("%d%c", (6 * i + d[j]) * k, j == 3 ? '\n' : ' ');
  }
}
int main() {
  clock_t tm = clock();
  solve();
  cerr << setprecision(3);
  cerr << (clock() - tm) / 1000000. << endl;
  return 0;
}