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 gcd(int x, int y) {
  if (x % y == 0) return y;
  return gcd(y, x % y);
}
int main() {
  int n, k, i, j;
  while (scanf("%d%d", &n, &k) != EOF) {
    printf("%d\n", (n * 6 - 1) * k);
    j = 0;
    for (i = 1; i <= n; i++) {
      printf("%d %d %d ", (i * 6 - 5) * k, (i * 6 - 3) * k, (i * 6 - 1) * k);
      for (j = j + 2;; j += 2) {
        if (gcd(i * 6 - 5, j) == 1 && gcd(i * 6 - 3, j) == 1 &&
            gcd(i * 6 - 1, j) == 1) {
          printf("%d\n", j * k);
          break;
        }
      }
    }
  }
  return 0;
}