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 std::cerr;
using std::cin;
using std::cout;
using std::deque;
using std::endl;
using std::fixed;
using std::ios_base;
using std::make_pair;
using std::map;
using std::max;
using std::min;
using std::ostream;
using std::pair;
using std::set;
using std::setprecision;
using std::string;
using std::stringstream;
using std::swap;
using std::vector;
using namespace std;
long long MOD = 1e9 + 7;
long long rev = 500000004;
int main() {
  ios_base::sync_with_stdio(0);
  long long n, k;
  cin >> n >> k;
  cout << k * (6 * n - 1) << '\n';
  for (long long i = 0; i < (long long)(n); ++i) {
    cout << k * (6 * i + 1) << " " << k * (6 * i + 3) << " " << k * (6 * i + 4)
         << " " << k * (6 * i + 5) << '\n';
  }
  return 0;
}