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 = 10000 + 10;
const int maxm = 2e5 + 10;
const long long mod = 1e9 + 7;
const int inf = 0x3ffffff;
const double eps = 1e-5;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long a[maxn][4];
long long n, k;
long long ans;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long p = 2, q = 1;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    a[i][0] = p;
    p += 2;
    if (p % 3 == 0) p += 2;
    a[i][1] = q;
    q += 2;
    a[i][2] = q;
    q += 2;
    a[i][3] = q;
    q += 2;
    for (int j = 0; j < 4; j++) {
      a[i][j] *= k;
      ans = max(ans, a[i][j]);
    }
  }
  cout << ans << endl;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 4; j++) cout << a[i][j] << " ";
    cout << endl;
  }
  return ~~(0 - 0);
}