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;
using ll = long long;
const int maxn = 1e4 + 10;
vector<ll> v[maxn];
int main(int argc, const char* argv[]) {
  ios::sync_with_stdio(false);
  ll n, k;
  cin >> n >> k;
  ll t = 1;
  for (int i = 0; i < n; ++i) {
    ll tt = t;
    for (; t < tt + 3; ++t) {
      v[i].push_back(t * k);
    }
    t++;
    v[i].push_back(t * k);
    t += 2;
  }
  cout << v[n - 1][3] << endl;
  for (int i = 0; i < n; ++i) {
    for (ll j : v[i]) {
      cout << j << " ";
    }
    cout << endl;
  }
  return 0;
}