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 int nextInt() {
  char ch;
  int x = 0;
  bool flag = false;
  do ch = (char)getchar(), flag = (ch == '-') ? true : flag;
  while (ch < '0' || ch > '9');
  do x = x * 10 + ch - '0';
  while (ch = (char)getchar(), ch <= '9' && ch >= '0');
  return x * (flag ? -1 : 1);
}
int n, m;
int gcd(int x, int y) { return (x % y == 0) ? y : gcd(y, x % y); }
set<int> S, S2;
int v[11000][4];
int main() {
  int i, j, k;
  int x, y, z;
  scanf("%d%d", &n, &k);
  for (i = 2; i <= 5000000; i += 2) {
    if (i % 3 == 0) continue;
    S2.insert(i);
  }
  bool flag;
  set<int>::iterator it1;
  int mx = 0;
  for (i = 0; i < n; i++) {
    v[i][0] = i * 6 + 1;
    v[i][1] = i * 6 + 3;
    v[i][2] = i * 6 + 5;
    for (it1 = S2.begin(); it1 != S2.end(); it1++) {
      flag = true;
      for (j = 0; j < 3; j++) {
        if (gcd(v[i][j], *it1) != 1) {
          flag = false;
          break;
        }
      }
      if (flag) {
        v[i][3] = *it1;
        S2.erase(it1);
        break;
      }
    }
    mx = v[i][2];
  }
  cout << mx * k << endl;
  for (i = 0; i < n; i++) {
    for (j = 0; j < 4; j++) {
      printf("%d ", v[i][j] * k);
    }
    printf("\n");
  }
  return 0;
}