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 long long int MOD = 1e9 + 7;
const double PI = 2 * acos(0.0);
void solve() {
  long long int n, k;
  cin >> n >> k;
  vector<long long int> v;
  v = {1, 3, 4, 5};
  long long int odd = 7;
  for (long long int i = 1; i < n; i++) {
    v.push_back(odd);
    v.push_back(odd + 2);
    v.push_back(odd + 3);
    v.push_back(odd + 4);
    odd += 6;
  }
  cout << (odd - 2) * k << endl;
  long long int ind = 0;
  for (long long int i = 0; i < n; i++) {
    cout << k * v[ind] << " " << k * v[ind + 1] << " " << k * v[ind + 2] << " "
         << k * v[ind + 3] << endl;
    ind += 4;
  }
}
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}