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 = 1E6 + 10;
int a[MAXN];
struct range {
  int bg, ed, step;
  range(int a, int b, int c) {
    bg = a;
    ed = b;
    step = c;
  }
  int next() { bg += step; }
  int end() { return bg > ed; }
  int now() { return bg; }
};
int getAns(long long n, long long k) {
  printf("%lld\n", (6 * n - 1) * k);
  for (long long i = 1; i <= n; ++i) {
    printf("%lld %lld %lld %lld\n", (6 * i - 5) * k, (6 * i - 4) * k,
           (6 * i - 3) * k, (6 * i - 1) * k);
  }
  return 0;
}
int main() {
  int n;
  int k;
  scanf("%d%d", &n, &k);
  getAns(n, k);
  return 0;
}