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 MOD = 1000000007;
template <class T>
inline T tmin(T a, T b) {
  return (a < b) ? a : b;
}
template <class T>
inline T tmax(T a, T b) {
  return (a > b) ? a : b;
}
template <class T>
inline void add_max(T &a, T b) {
  if (b > a) a = b;
}
template <class T>
inline void add_min(T &a, T b) {
  if (b < a) a = b;
}
template <class T>
inline T tabs(T a) {
  return (a > 0) ? a : -a;
}
template <class T>
T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main(int argc, char *argv[]) {
  ios_base::sync_with_stdio(false);
  int n, k, ans;
  cin >> n >> k;
  ans = k * (6 * n - 1);
  printf("%d\n", ans);
  int t = 1;
  for (int i = (0); i < (n); ++i) {
    int s = (t << 1) * k;
    printf("%d %d %d %d\n", s - k, s, s + k, s + 3 * k);
    t += 3;
  }
  return 0;
}