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>
int n, m, ans;
inline long long read() {
  long long x = 0;
  bool flag = 0;
  char ch = getchar();
  while (!isdigit(ch) && ch != '-') ch = getchar();
  if (ch == '-') flag = 1, ch = getchar();
  while (isdigit(ch)) x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
  if (flag) x = -x;
  return x;
}
struct chain {
  int l, r, v;
} a[200005];
bool vis[200005];
std::priority_queue<std::pair<int, int> > q;
int main() {
  n = read();
  m = read();
  int now = 0;
  printf("%d\n", m * ((n - 1) * 6 + 5));
  for (int i = (1); i <= (n); ++i) {
    printf("%d %d %d %d\n", m * (now + 1), m * (now + 2), m * (now + 3),
           m * (now + 5));
    now += 6;
  }
  return 0;
}