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;
void file(string s) {
  freopen((s + ".in").c_str(), "r", stdin);
  freopen((s + ".out").c_str(), "w", stdout);
}
template <typename Tp>
void read(Tp &x) {
  int fh = 1;
  char c = getchar();
  x = 0;
  while (c > '9' || c < '0') {
    if (c == '-') {
      fh = -1;
    }
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + (c & 15);
    c = getchar();
  }
  x *= fh;
}
int n, m;
int ans[1000005][4];
void make(int id) {
  int x = 1;
  if (id > 1) x = ans[id - 1][3] + 2;
  if (x % 3 == 2) x += 2;
  ans[id][0] = x;
  ans[id][1] = x + 1;
  ans[id][2] = x + 2;
  ans[id][3] = x + 4;
}
signed main() {
  read(n);
  read(m);
  for (int i = 1; i <= n; ++i) {
    make(i);
  }
  printf("%lld\n", 1ll * ans[n][3] * m);
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < 4; ++j) {
      printf("%lld%c", 1ll * ans[i][j] * m, " \n"[j == 3]);
    }
  }
  return 0;
}