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;
bool cmp(int a, int b) { return a > b; }
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <class T>
T lcm(T a, T b) {
  return a * b / gcd(a, b);
}
long long n, k;
vector<long long> ans[100005];
long long maxx(long long a, long long b) {
  if (a > b) return a;
  return b;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k;
  long long maa = k;
  for (int i = 0; i < n; i++) {
    int a1 = (6 * i + 1) * k, a2 = (6 * i + 2) * k, a3 = (6 * i + 3) * k,
        a4 = (6 * i + 5) * k;
    maa = maxx(maa, a1);
    maa = maxx(maa, a2);
    maa = maxx(maa, a3);
    maa = maxx(maa, a4);
    ans[i].push_back(a1);
    ans[i].push_back(a2);
    ans[i].push_back(a3);
    ans[i].push_back(a4);
  }
  cout << maa << endl;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 4; j++) cout << ans[i][j] << " ";
    cout << endl;
  }
  return 0;
}