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 A();
void B();
void C();
void D();
int main() {
  D();
  return 0;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
void D() {
  int n, k;
  cin >> n >> k;
  int x = 7;
  int arr[n][4];
  arr[0][0] = k;
  arr[0][1] = 2 * k;
  arr[0][2] = 3 * k;
  arr[0][3] = 5 * k;
  for (int i = 1; i < n; i++) {
    arr[i][1] = x * k;
    x += 2;
    arr[i][2] = x * k;
    x += 2;
    arr[i][3] = x * k;
    x += 2;
    arr[i][0] = arr[i][3] - k;
  }
  printf("%d\n", arr[n - 1][3]);
  for (int i = 0; i < n; i++) {
    printf("%d %d %d %d\n", arr[i][0], arr[i][1], arr[i][2], arr[i][3]);
  }
}