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;
long long arr[10000][4];
long long gcd(long long a, long long b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
int main() {
  int n, k;
  cin >> n >> k;
  long long odd = 1, even = 2;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 3; ++j) {
      arr[i][j] = odd;
      odd += 2;
    }
    while (1) {
      bool ok = 1;
      for (int j = 0; j < 3; ++j)
        if (gcd(arr[i][j], even) != 1) ok = 0;
      if (ok) break;
      even += 2;
    }
    arr[i][3] = even;
    even += 2;
  }
  cout << (odd - 2) * k << endl;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 4; ++j) printf("%I64d ", arr[i][j] * k);
    printf("\n");
  }
  return 0;
}