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;
template <class T>
inline T sqr(T x) {
  return x * x;
}
const int inf = ~0u >> 1;
const double eps = 1e-8;
const double pi = acos(-1.0);
int sgn(double x) { return fabs(x) < eps ? 0 : x < -eps ? -1 : 1; }
int gcd(int a, int b) { return (b == 0) ? a : gcd(b, a % b); }
int main() {
  int n, k;
  cin >> n >> k;
  int cnt = 0;
  vector<vector<int> > biu;
  biu.resize(n);
  vector<int> temp = vector<int>{1, 2, 3, 5};
  biu[0] = temp;
  int even_idx = 4, odd_idx = 7;
  for (int i = 1; i < n; i++) {
    temp[1] = odd_idx;
    temp[2] = odd_idx += 2;
    temp[3] = odd_idx += 2;
    odd_idx += 2;
    bool flag = true;
    for (int j = 1; j <= 3; j++) {
      if (gcd(even_idx, temp[j]) != 1) {
        flag = false;
        break;
      }
    }
    if (flag) {
      temp[0] = even_idx;
    } else {
      even_idx += 2;
      temp[0] = even_idx;
    }
    even_idx += 2;
    biu[i] = temp;
  }
  cout << biu[n - 1][3] * k << endl;
  for (auto v : biu) {
    for (auto i : v) {
      cout << i * k << ' ';
    }
    cout << endl;
  }
}