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;
int conversion(string p) {
  int o;
  o = atoi(p.c_str());
  return o;
}
string toString(int h) {
  stringstream ss;
  ss << h;
  return ss.str();
}
string toStringLong(long long 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)); }
int lcm(int a, int b) { return (a * (b / gcd(a, b))); }
int main() {
  int n, k;
  cin >> n >> k;
  int tam[4];
  memset(tam, 0, sizeof tam);
  tam[0] = k;
  cout << 5 * k + (6 * k) * (n - 1) << endl;
  for (int i = 0; i < n; i++) {
    for (int j = 1; j < 4; j++) {
      if (j == 1) {
        tam[j] = tam[j - 1] + k * 2;
      } else {
        tam[j] = tam[j - 1] + k;
      }
    }
    cout << tam[0] << " " << tam[1] << " " << tam[2] << " " << tam[3] << endl;
    tam[0] = tam[3] + k * 2;
  }
}