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 = 1e9 + 7;
const int maxn = 10000;
int n, k;
int a[maxn + 5][4];
int ans;
void init() { scanf("%d%d", &n, &k); }
int gcd(int x, int y) {
  while (y) y ^= x ^= y ^= x %= y;
  return x;
}
void solve() {
  memset(a, -1, sizeof a);
  for (int i = (0), _end_ = (n); i != _end_; ++i)
    a[i][0] = 6 * i + 1, a[i][1] = 6 * i + 2, a[i][2] = 6 * i + 3,
    a[i][3] = 6 * i + 5, ans = a[i][3] * k;
}
void output() {
  printf("%d\n", ans);
  for (int i = (0), _end_ = (n); i != _end_; ++i)
    printf("%d %d %d %d\n", a[i][0] * k, a[i][1] * k, a[i][2] * k, a[i][3] * k);
}
int main() {
  init();
  solve();
  output();
  return 0;
}