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;
long long gcd(long long a, long long b) {
  if (!a) return b;
  if (!b) return a;
  return gcd(b, a % b);
}
int n, k;
long long next(long long prev, long long a, long long b, long long c) {
  prev += 2;
  while (gcd(prev * k, a * k) != k || gcd(prev * k, b * k) != k ||
         gcd(prev * k, c * k) != k)
    prev += 2;
  return prev;
}
int main() {
  scanf("%d %d", &n, &k);
  long long a2 = 1;
  printf("%d\n", (6 * (n - 1) + 5) * k);
  long long p = 0;
  for (long long i = (0); i < (n); ++i) {
    p = next(p, a2, a2 + 2, a2 + 4);
    printf("%lld %lld %lld %lld\n", p * k, a2 * k, (a2 + 2) * k, (a2 + 4) * k);
    a2 += 6;
  }
  return 0;
}