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>
void read(T& num) {
  char CH;
  bool F = false;
  for (CH = getchar(); CH < '0' || CH > '9'; F = CH == '-', CH = getchar())
    ;
  for (num = 0; CH >= '0' && CH <= '9';
       num = num * 10 + CH - '0', CH = getchar())
    ;
  F && (num = -num);
}
int stk[70], tp;
template <class T>
inline void print(T p) {
  if (!p) {
    puts("0");
    return;
  }
  while (p) stk[++tp] = p % 10, p /= 10;
  while (tp) putchar(stk[tp--] + '0');
  putchar('\n');
}
const int mod = 1e9 + 7;
const double PI = acos(-1.0);
const int inf = 1e9;
const int N = 1e6 + 20;
const int maxn = 1e5 + 110;
const double eps = 1e-12;
int main() {
  int n, k;
  read(n);
  read(k);
  cout << (n - 1) * 6 * k + 5 * k << endl;
  for (int i = 1; i <= n; i++) {
    int t = (i - 1) * 6 + 1;
    printf("%d %d %d %d\n", t * k, t * k + k, t * k + 2 * k, t * k + 4 * k);
  }
  return 0;
}