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 int MAXN = 100000 + 2;
int gcd(int a, int b) {
  int r = a % b;
  while (r > 0) {
    a = b;
    b = r;
    r = a % b;
  }
  return (b);
}
int main() {
  int n, k;
  cin >> n >> k;
  vector<vector<int> > a(n);
  int x = 0, y = -1;
  for (int i = 0; i < n; i++) {
    y += 2;
    a[i].push_back(y);
    y += 2;
    a[i].push_back(y);
    y += 2;
    a[i].push_back(y);
    while (true) {
      x += 2;
      bool ok = true;
      for (int j = 0; j < a[i].size(); j++) {
        if (gcd(a[i][j], x) > 1) {
          ok = false;
          break;
        }
      }
      if (ok) break;
    }
    a[i].push_back(x);
  }
  cout << (long long)(max(x, y)) * k << endl;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < a[i].size(); j++)
      cout << (long long)(a[i][j]) * k << " ";
    cout << endl;
  }
}