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;
inline long long in() {
  int32_t x;
  scanf("%d", &x);
  return x;
}
inline long long lin() {
  long long x;
  scanf("%lld", &x);
  return x;
}
char ch[4000010];
inline string get() {
  scanf("%s", ch);
  return string(ch);
}
inline void read(long long *a, long long n) {
  for (long long i = 0; i < n; i++) a[i] = in();
}
template <class blank>
inline void out(blank x) {
  cout << x << "\n";
  exit(0);
}
template <class blank, class blank2>
inline void smin(blank &a, blank2 b) {
  a = a <= b ? a : b;
}
template <class blank, class blank2>
inline void smax(blank &a, blank2 b) {
  a = a >= b ? a : b;
}
const long long maxn = 5e5 + 10;
const long long maxm = 4e6 + 10;
const long long maxlg = 20;
const long long base = 29;
const long long mod = 1e9 + 7;
const long long inf = 2e18 + 10;
const double eps = 1e-9;
int32_t main() {
  long long n = in(), k = in();
  cout << 6 * n * k - k << "\n";
  for (long long i = 0; i < n; i++) {
    printf("%lld %lld %lld %lld\n", (6 * i + 1) * k, (6 * i + 2) * k,
           (6 * i + 3) * k, (6 * i + 5) * k);
  }
}