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;
vector<int> ans[10005];
int gcd(int x, int y) {
  int z;
  while (y) {
    z = x;
    x = y;
    y = z % y;
  }
  return x;
}
int main() {
  int n, k, cnt = 0, sol = 0;
  scanf("%d%d", &n, &k);
  for (int i = 1, j = 2; cnt < n; j += 2)
    if (j % 3)
      if (gcd(i, j) == 1 && gcd(i + 2, j) == 1 && gcd(i + 4, j) == 1) {
        ans[++cnt].push_back(i * k);
        ans[cnt].push_back((i + 2) * k);
        ans[cnt].push_back((i + 4) * k);
        ans[cnt].push_back(j * k);
        i += 6;
        sol = max((i - 2) * k, j * k);
      }
  printf("%d\n", sol);
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < 4; ++j) printf("%d ", ans[i][j]);
    printf("\n");
  }
  return 0;
}