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;
template <class T>
inline T bigmod(T p, T e, T M) {
  long long ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return (T)ret;
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
int ar[10001][4];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long a, b = 0, c, d = 0, e, f = 0, g, h = 0, x, y, z = 0;
  cin >> a >> b;
  c = 1;
  for (int i = (1); i < (a + 1); ++i) {
    ar[i][0] = c;
    ar[i][1] = c + 1;
    ar[i][2] = c + 2;
    ar[i][3] = c + 4;
    if (i == a) d = ar[i][3];
    c += 6;
  }
  cout << d * b << endl;
  for (int i = (1); i < (a + 1); ++i) {
    for (int j = (0); j < (4); ++j) cout << ar[i][j] * b << " ";
    cout << endl;
  }
}