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 gcd(int a, int b) {
  while (a) {
    swap(b %= a, a);
  }
  return b;
}
int n, k;
struct st {
  int el[4];
};
bool test(const st& el, int c) {
  for (int i = 0; i < c; i++) {
    if (gcd(el.el[i], el.el[c]) != 1) return false;
  }
  return true;
}
int main() {
  cin >> n >> k;
  list<int> tmp;
  int mx = 1;
  vector<st> res;
  int cur = 1;
  int last = 2;
  for (int step = 1; step <= n; ++step) {
    st el;
    el.el[0] = cur;
    cur += 2;
    el.el[1] = cur;
    cur += 2;
    el.el[2] = cur;
    mx = cur;
    cur += 2;
    while (true) {
      el.el[3] = (last);
      last += 2;
      if (test(el, 3)) {
        mx = max(mx, el.el[3]);
        break;
      }
    }
    res.push_back(el);
  }
  cout << k * mx << '\n';
  for (int i = 0; i < res.size(); i++) {
    for (int j = 0; j < 4; j++) cout << res[i].el[j] * k << ' ';
    cout << '\n';
  }
  return 0;
}