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;
int n, k, x[100000][5], len, now;
void print(int k1) {
  printf("%d\n", k1 * k);
  for (int i = 1; i <= len; i++) {
    if (x[i][0] == 4)
      printf("%d %d %d %d\n", x[i][1] * k, x[i][2] * k, x[i][3] * k,
             x[i][4] * k);
  }
}
int gcd(int k1, int k2) {
  if (k2 == 0) return k1;
  return gcd(k2, k1 % k2);
}
int main() {
  scanf("%d%d", &n, &k);
  printf("%d\n", (6 * (n - 1) + 5) * k);
  for (int i = 0; i < n; i++)
    printf("%d %d %d %d\n", (6 * i + 1) * k, (6 * i + 2) * k, (6 * i + 3) * k,
           (6 * i + 5) * k);
  return 0;
}