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 a, int b) { return b ? gcd(b, a % b) : a; }
bool coprime(int a, int b) { return gcd(a, b) == 1; }
int ans[10005][4];
int main() {
  int N, K;
  cin >> N >> K;
  int mx = 1;
  int i = 1;
  for (int x = 0; x < (N); ++x) {
    ans[x][0] = i;
    ans[x][1] = i + 1;
    ans[x][2] = i + 2;
    ans[x][3] = i + 4;
    mx = i + 4;
    i += 6;
  }
  printf("%d\n", mx * K);
  for (int x = 0; x < (N); ++x) {
    printf("%d %d %d %d\n", K * ans[x][0], K * ans[x][1], K * ans[x][2],
           K * ans[x][3]);
  }
}