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 n, k;
int a[10010][5];
int gcd(int x, int y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
int main() {
  scanf("%d%d", &n, &k);
  a[1][1] = 1;
  a[1][2] = 2;
  a[1][3] = 3;
  a[1][4] = 5;
  a[2][1] = 4;
  a[2][2] = 7;
  a[2][3] = 9;
  a[2][4] = 11;
  for (int i = 3; i <= n; i++) {
    a[i][4] = 6 * i - 1;
    a[i][3] = 6 * i - 3;
    a[i][2] = 6 * i - 5;
    int t = a[i - 1][1] + 2;
    for (int j = 0; t + 2 * j < a[i][4]; j++) {
      if (gcd(t + 2 * j, a[i][2]) == 1 && gcd(t + 2 * j, a[i][3]) == 1 &&
          gcd(t + 2 * j, a[i][4]) == 1) {
        a[i][1] = t + 2 * j;
        break;
      }
    }
  }
  printf("%d\n", k * a[n][4]);
  for (int i = 1; i <= n; i++)
    printf("%d %d %d %d\n", k * a[i][1], k * a[i][2], k * a[i][3], k * a[i][4]);
}