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;
int z[10000000], n, k;
int gcd(int x, int y) {
  if (x == 0) return y;
  return gcd(y % x, x);
}
int main() {
  int n;
  cin >> n >> k;
  vector<pair<pair<int, int>, pair<int, int>>> ans(n);
  int p = -1;
  for (int(i) = (0); i < (n); ++(i)) {
    ans[i].first.second = p + 2;
    p += 2;
    ans[i].second.first = p + 2;
    p += 2;
    ans[i].second.second = p + 2;
    p += 2;
  }
  int q = 2;
  for (int(i) = (0); i < (n); ++(i)) {
    while (gcd(q, ans[i].second.second) != 1 ||
           gcd(q, ans[i].first.second) != 1 || gcd(q, ans[i].second.first) != 1)
      q += 2;
    ans[i].first.first = q;
    q += 2;
  }
  cout << k * p << "\n";
  for (int(i) = (0); i < (n); ++(i))
    cout << ans[i].first.first * k << " " << ans[i].first.second * k << " "
         << ans[i].second.first * k << " " << ans[i].second.second * k << "\n";
  return 0;
}