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;
int bs(int *st, int *en, int val, char c) {
  int k;
  if (c == 'l') k = lower_bound(st, en, val) - st;
  if (c == 'u') k = upper_bound(st, en, val) - st;
  return k;
}
void ai(int *p, int l) {
  for (int a = 0; a < l; a++) scanf("%d", &p[a]);
}
void al(int *p, int l) {
  for (int a = 0; a < l; a++) cin >> p[a];
}
int prime[78500];
int sieve() {
  int a, b, c;
  c = 0;
  prime[c] = 2;
  bool *m = (bool *)calloc(1000006, sizeof(bool));
  for (a = 3; a <= 1000000; a = a + 2) {
    if (!m[a]) {
      prime[++c] = a;
      for (b = 2 * a; b <= 1000000; b = b + a) m[b] = true;
    }
  }
  free(m);
  return c;
}
vector<long long> v[200];
int main() {
  long long a, b, c, e, z, x, y, s;
  long long n, k;
  cin >> n >> k;
  e = 1;
  for (a = 0;; a++) {
    x = 3 * a + 1;
    if ((x % 2)) {
      v[e].push_back(k * x);
      v[e].push_back(k * (x + 1));
      v[e].push_back(k * (x + 2));
      v[e].push_back(k * (x + 4));
      e++;
    }
    if (e > n) break;
  }
  cout << v[n][3] << endl;
  for (a = 1; a <= n; a++)
    cout << v[a][0] << " " << v[a][1] << " " << v[a][2] << " " << v[a][3]
         << endl;
  return 0;
}