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 MOD = 1e9 + 7;
const int MAXN = 1e5 + 5;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int main(void) {
  int n, k;
  cin >> n >> k;
  int quad[n][4];
  int p = 1;
  for (int i = 0; i < n; i++) {
    if (p % 2 == 0) p++;
    quad[i][0] = p++;
    int prev = p;
    quad[i][1] = p++;
    quad[i][2] = p++;
    while (gcd(p, prev) > 1 || gcd(p, prev + 1) > 1 || gcd(p, prev - 1) > 1)
      p++;
    quad[i][3] = p++;
  }
  cout << 1LL * quad[n - 1][3] * k << '\n';
  for (int i = 0; i < n; i++)
    cout << 1LL * quad[i][0] * k << ' ' << 1LL * quad[i][1] * k << ' '
         << 1LL * quad[i][2] * k << ' ' << 1LL * quad[i][3] * k << '\n';
  return 0;
}