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 double eps = 1e-8;
const double PI = acos(-1.);
const long long MOD = 1000000007;
vector<int> G[10005];
int main() {
  int i, j, k, _T = 0;
  int n, m;
  while (~scanf("%d%d", &n, &m)) {
    for (i = 1; i <= n; i++) G[i].clear();
    G[1].push_back(1);
    G[1].push_back(2);
    G[1].push_back(3);
    G[1].push_back(5);
    int j = 4;
    for (i = 2; i <= n; i++) {
      G[i].push_back(j), G[i].push_back(6 * i - 5), G[i].push_back(6 * i - 3),
          G[i].push_back(6 * i - 1);
      j += 2;
      if (j % 3 == 0) j += 2;
    }
    int f = 0;
    for (i = 1; i <= n; i++)
      for (j = 0; j < G[i].size(); j++) f = max(f, G[i][j]);
    cout << f * m << "\n";
    for (i = 1; i <= n; i++, cout << "\n")
      for (j = 0; j < G[i].size(); j++) cout << G[i][j] * m << " ";
  }
  return 0;
}