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;
set<pair<int, int> > S;
int N, K;
const int limit = 600005;
bool V[limit];
int Prime[limit];
long long ans = 5;
vector<int> Sol[10005];
map<int, int> M;
void Eratostenes() {
  for (int i = 2; i <= limit; i++) {
    if (V[i] == 0) {
      Prime[++Prime[0]] = i;
      for (int j = i + i; j <= limit; j += i) V[j] = 1;
    }
  }
}
void Solve() {
  Sol[1].push_back(1);
  Sol[1].push_back(2);
  Sol[1].push_back(3);
  Sol[1].push_back(5);
  S.insert(make_pair(4, 2));
  S.insert(make_pair(9, 3));
  S.insert(make_pair(25, 5));
  for (int i = 4; i <= Prime[0]; i++) {
    S.insert(make_pair(Prime[i], Prime[i]));
  }
  for (int i = 2; i <= N; i++) {
    auto it = S.begin();
    int cnt = 0;
    while (cnt < 4) {
      it = S.begin();
      while (M[it->second] != 0) it = next(it);
      ++cnt;
      Sol[i].push_back(it->first);
      long long a = it->first, b = it->second;
      ans = max(ans, a);
      S.erase(it);
      S.insert(make_pair(a * b, b));
      M[b] = 1;
    }
    M.clear();
  }
  ans *= K;
  cout << ans << "\n";
  for (int i = 1; i <= N; i++) {
    for (int j = 0; j < 4; j++) cout << Sol[i][j] * K << " ";
    cout << "\n";
  }
}
int main() {
  cin >> N >> K;
  cout << (6 * N - 1) * K << "\n";
  for (int i = 0; i < N; i++) {
    cout << (6 * i + 1) * K << " ";
    cout << (6 * i + 2) * K << " ";
    cout << (6 * i + 3) * K << " ";
    cout << (6 * i + 5) * K << "\n";
  }
  return 0;
}