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 maxn = 4e6 + 5;
int n, k;
int prime[maxn], pl;
bool isprime[maxn];
void make_prime() {
  int i, j;
  pl = 0;
  fill(isprime, isprime + maxn, true);
  prime[pl++] = 1;
  for (i = 2; i < maxn; i++) {
    if (isprime[i]) {
      prime[pl++] = i;
      for (j = 2 * i; j < maxn; j += i) isprime[j] = false;
    }
  }
}
int main() {
  int i, j;
  make_prime();
  scanf("%d%d", &n, &k);
  printf("%d\n", (6 * (n - 1) + 5) * k);
  for (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;
}