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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
long long i = 0, j = 0, k = 0, cnt = 0, sum = 0, res = 0, start = 0, finall = 0,
          maxi = 0, mini = 1000000007, len = 0, cnt1 = 0, cnt2 = 0, ans = 0,
          cnt3 = 0, eend = 0, pos = 0;
int flag = 0;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, k;
  cin >> n >> k;
  cout << (6 * (n - 1) + 5) * k << "\n";
  for (i = 1; i <= n; i++) {
    cout << (6 * (i - 1) + 1) * k << " ";
    cout << (6 * (i - 1) + 2) * k << " ";
    cout << (6 * (i - 1) + 3) * k << " ";
    cout << (6 * (i - 1) + 5) * k << "\n";
  }
}