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;
int64_t mod = 1e9 + 7;
int64_t mod1 = 1e9 + 5;
int64_t power(int64_t a, int64_t b) {
  if (b == 0)
    return 1;
  else if (b % 2 == 0)
    return power((((a % mod) * (a % mod)) % mod), b / 2) % mod;
  else
    return ((a % mod) * (power((((a % mod) * (a % mod)) % mod), b / 2) % mod)) %
           mod;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int64_t n, k;
  cin >> n >> k;
  int64_t num = n - 1;
  cout << (6 * num + 5) * k << '\n';
  for (int64_t i = 0; i < n; i++) {
    cout << (6 * i + 1) * k << " " << (6 * i + 3) * k << " " << (6 * i + 2) * k
         << " " << (6 * i + 5) * k << " ";
    cout << '\n';
  }
}