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 bool getmin(T *a, const T &b) {
  if (b < *a) {
    *a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool getmax(T *a, const T &b) {
  if (b > *a) {
    *a = b;
    return true;
  }
  return false;
}
template <class T>
inline void read(T *a) {
  char c;
  while (isspace(c = getchar())) {
  }
  bool flag = 0;
  if (c == '-')
    flag = 1, *a = 0;
  else
    *a = c - 48;
  while (isdigit(c = getchar())) *a = *a * 10 + c - 48;
  if (flag) *a = -*a;
}
const int mo = 1000000007;
template <class T>
T pow(T a, T b, int c = mo) {
  T res = 1;
  for (T i = 1; i <= b; i <<= 1, a = 1LL * a * a % c)
    if (b & i) res = 1LL * res * a % c;
  return res;
}
int n, k;
int main() {
  cin >> n >> k;
  cout << (6 * n - 1) * k << endl;
  for (int i = (0); i <= (n - 1); ++i)
    printf("%d %d %d %d\n", (6 * i + 1) * k, (6 * i + 2) * k, (6 * i + 3) * k,
           (6 * i + 5) * k);
  return 0;
}