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>
int gcd(int a, int b) {
  if (a < b) std::swap(a, b);
  while (0 < b) {
    a %= b;
    std::swap(a, b);
  }
  return a;
}
int main() {
  int n, k;
  int e(2), o(1), m(0);
  int w[3];
  std::vector<int> r;
  std::scanf("%d%d", &n, &k);
  for (int i(0); i < n; ++i) {
    for (int j(0); j < 3; ++j) {
      r.push_back(o + j * 2);
      w[j] = o + j * 2;
    }
    o += 6;
    m = w[2];
  N:
    for (int j(0); j < 3; ++j) {
      if (gcd(e, w[j]) != 1) {
        e += 2;
        goto N;
      }
    }
    r.push_back(e);
    e += 2;
  }
  std::printf("%d\n", m * k);
  for (int i(0); i < n; ++i) {
    for (int j(0); j < 4; ++j) {
      if (0 < j) std::printf(" ");
      std::printf("%d", r[i * 4 + j] * k);
    }
    std::puts("");
  }
  return 0;
}