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>
#pragma comment(linker, "/STACK:100000000")
using namespace std;
const bool db = false;
int n, k;
vector<int> v;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
int main() {
  scanf("%d%d", &n, &k);
  int cur = 1;
  for (int i = 0; i < n; ++i) {
    v.push_back(cur);
    v.push_back(cur + 1);
    v.push_back(cur + 2);
    v.push_back(cur + 4);
    cur += 6;
  }
  cout << v.back() * k << endl;
  for (int i = 0; i < n; ++i) {
    for (int j = 4 * i; j < 4 * i + 4; ++j) {
      cout << v[j] * k << " ";
    }
    cout << endl;
  }
  cin.get();
  cin.get();
  return 0;
}