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;
long long toi(string s) {
  istringstream is(s);
  long long x;
  is >> x;
  return x;
}
int conversion(string p) {
  int num;
  num = atoi(p.c_str());
  return num;
}
string toString(int h) {
  stringstream ss;
  ss << h;
  return ss.str();
}
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
long long lcm(long long a, long long b) { return (a * (b / gcd(a, b))); }
bool cmp(pair<int, int> a, pair<int, int> b) { return a.first > b.first; }
void solve() {
  long long n, k;
  cin >> n >> k;
  cout << ((6 * n) - 1) * k << endl;
  for (int i = 0; i < n; i++) {
    cout << ((6 * i) + 1) * k << " " << ((6 * i) + 2) * k << " "
         << ((6 * i) + 3) * k << " " << ((6 * i) + 5) * k << endl;
  }
}
int main() {
  solve();
  return 0;
}