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 Node {
  long long a, b, c, d;
  Node(long long a = 0, long long b = 0, long long c = 0, long long d = 0)
      : a(a), b(b), c(c), d(d) {}
  void outpt(long long k) {
    printf("%I64d %I64d %I64d %I64d\n", a * k, b * k, c * k, d * k);
  }
};
vector<Node> ans;
int main() {
  long long n, k, m = 1;
  while (cin >> n >> k) {
    for (int i = 1; i <= n; i++) {
      ans.push_back(Node(m, m + 1, m + 2, m + 4));
      m += 6;
    }
    m -= 2;
    cout << m * k << endl;
    for (int i = 0; i < n; i++) ans[i].outpt(k);
  }
  return 0;
}