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 read() {
  long long sum = 0, f = 0;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar()) f |= (ch == '-');
  for (; isdigit(ch); ch = getchar()) sum = (sum << 1) + (sum << 3) + (ch ^ 48);
  return f ? -sum : sum;
}
void write(const long long x) {
  if (x < 0) {
    putchar('-');
    write(-x);
    return;
  }
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
int n, k, x;
int main() {
  n = read(), k = read();
  write((n * 6 - 1) * k), puts("");
  for (register long long i = (1); i <= (n); i++)
    x = i * 6 - 5, write(x * k), putchar(' '), write((x + 1) * k), putchar(' '),
    write((x + 2) * k), putchar(' '), write((x + 4) * k), puts("");
}