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;
vector<int> ans, tmp;
int gcd(int x, int y) { return x ? gcd(y % x, x) : y; }
int main() {
  int N, K;
  scanf("%d%d", &N, &K);
  for (int i = 1;; i++) {
    bool ok = true;
    if (tmp.size() == 0 && i % 3 == 0) continue;
    for (int j = tmp.size() - 1; j >= 0; j--)
      if (gcd(i, tmp[j]) != 1) ok = false;
    if (ok == true) tmp.push_back(i);
    if (tmp.size() == 4) {
      for (int _ = 0; _ < 4; _++) ans.push_back(tmp[_]);
      tmp.clear();
    }
    if (ans.size() == 4 * N) break;
  }
  printf("%I64d\n", K * 1LL * ans.back());
  for (int i = 0; i < N * 4; i++) {
    for (int _ = 0; _ < 4; _++)
      printf("%I64d%c", K * 1LL * ans[i + _], " \n"[_ == 3]);
    i += 3;
  }
  return 0;
}