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;
const int MAXN = 511111;
const int MOD = 1e9 + 7;
const double eps = 1e-6;
const long long MAXL = (long long)(0x7fffffffffffffff);
const int MAXI = 0x7fffffff;
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  int n, k;
  while (cin >> n >> k) {
    vector<int> nums(n * 4);
    int x = 2, y = 1;
    int mx = 0;
    for (int i = 0; i < n; i++) {
      for (int j = 1; j < 4; j++) {
        nums[i * 4 + j] = y * k;
        mx = max(y * k, mx);
        y += 2;
      }
      while (1) {
        bool can = true;
        for (int j = 1; j < 4 && can; j++) {
          if (gcd(x, nums[i * 4 + j] / k) != 1) can = false;
        }
        if (can) break;
        x += 2;
      }
      nums[i * 4] = x * k;
      mx = max(x * k, mx);
      x += 2;
    }
    cout << mx << endl;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < 4; j++) {
        cout << nums[i * 4 + j] << " ";
      }
      cout << endl;
    }
  }
}