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 res[10004][4];
int gcd(int a, int b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
int main() {
  int n = 0, k;
  scanf("%d%d", &n, &k);
  int mx = 1, m2 = 0;
  for (int t = 0; t < n; t++) {
    res[t][0] = mx;
    mx += 2;
    res[t][1] = mx;
    mx += 2;
    res[t][2] = mx;
    mx += 2;
    if (gcd(res[t][0], res[t][0] + 3) == 1 &&
        gcd(res[t][1], res[t][0] + 3) == 1 &&
        gcd(res[t][2], res[t][0] + 3) == 1) {
      res[t][3] = res[t][0] + 1;
    } else if (gcd(res[t][0], res[t][0] + 3) == 1 &&
               gcd(res[t][1], res[t][0] + 3) == 1 &&
               gcd(res[t][2], res[t][0] + 3) == 1) {
      res[t][3] = res[t][0] + 3;
    }
  }
  mx = max(mx - 2, m2);
  printf("%d\n", mx * k);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 4; j++) printf("%d ", res[i][j] * k);
    printf("\n");
  }
  return 0;
}