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 gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
bool Check(int x, int y) {
  if (gcd(x, y) > 1) return false;
  if (gcd(x + 1, y) > 1) return false;
  if (gcd(x + 2, y) > 1) return false;
  return true;
}
vector<int> ans[10010];
int main() {
  int n, k;
  cin >> n >> k;
  int x = 1;
  for (int i = 0; i < (n); i++) {
    ans[i].push_back(x * k);
    ans[i].push_back((x + 1) * k);
    ans[i].push_back((x + 2) * k);
    int y = x + 4;
    while (!Check(x, y)) y += 2;
    ans[i].push_back(y * k);
    x = y + 2;
  }
  cout << (x - 2) * k << endl;
  for (int i = 0; i < (n); i++)
    cout << ans[i][0] << " " << ans[i][1] << " " << ans[i][2] << " "
         << ans[i][3] << endl;
  return 0;
}