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;
struct data {
  int a, b, c, d;
  data() {}
  data(int a, int b, int c, int d) : a(a), b(b), c(c), d(d) {}
};
const int MAX = 100500;
data info[MAX];
int main() {
  std::ios_base::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  info[0] = data(1, 2, 3, 5);
  int mx;
  for (int i = 1; i < n; i++) {
    info[i] = data(info[i - 1].a + 6, info[i - 1].b + 6, info[i - 1].c + 6,
                   info[i - 1].d + 6);
  }
  mx = info[n - 1].d * k;
  cout << mx << "\n";
  for (int i = 0; i < n; i++) {
    cout << info[i].a * k << " " << info[i].b * k << " " << info[i].c * k << " "
         << info[i].d * k << "\n";
  }
  return 0;
}