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 MAX = 1e5 + 1;
const int MOD = 1000000007;
int dr[] = {0, -1, 0, 1};
int dc[] = {1, 0, -1, 0};
inline int Pow(int b, int p) {
  if (!p) return 1;
  int sq = Pow(b, p >> 1);
  sq *= sq;
  if (p & 1) sq *= b;
  return sq;
}
inline int gcd(int a, int b) {
  if (!a) return b;
  return gcd(b % a, a);
}
int main() {
  int n, k;
  while (cin >> n >> k) {
    int x = n - 1;
    printf("%d\n", k * (6 * x + 5));
    for (int i = 0; i < n; i++)
      printf("%d %d %d %d\n", k * (6 * i + 1), k * (6 * i + 2), k * (6 * i + 3),
             k * (6 * i + 5));
  }
  return 0;
}