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>
const double EPS = 1e-24;
const long long int MOD = 1000000007ll;
const double PI = 3.14159265359;
int INF = 2147483645;
template <class T>
T Max2(T a, T b) {
  return a < b ? b : a;
}
template <class T>
T Min2(T a, T b) {
  return a < b ? a : b;
}
template <class T>
T Max3(T a, T b, T c) {
  return Max2(Max2(a, b), c);
}
template <class T>
T Min3(T a, T b, T c) {
  return Min2(Min2(a, b), c);
}
template <class T>
T Max4(T a, T b, T c, T d) {
  return Max2(Max2(a, b), Max2(c, d));
}
template <class T>
T Min4(T a, T b, T c, T d) {
  return Min2(Min2(a, b), Max2(c, d));
}
template <class T>
void Swap(T& a, T& b) {
  T temp;
  temp = a;
  a = b;
  b = temp;
}
using namespace std;
int N, K;
int main() {
  std::ios::sync_with_stdio(false);
  cin >> N >> K;
  cout << (6 * N - 1) * K << "\n";
  for (int i = 0; i < N; i++)
    cout << K * (6 * i + 1) << " " << K * (6 * i + 2) << " " << K * (6 * i + 3)
         << " " << K * (6 * i + 5) << "\n";
  return 0;
}