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;
struct node {
  int ou, j1, j2, j3;
} a[10020];
void setup() {
  int i, j;
  int ou = 2, ji = 1;
  for (i = 1; i <= 10010; i++) {
    while (ou % 3 == 0) ou += 2;
    a[i].ou = ou;
    ou += 2;
    a[i].j1 = ji;
    a[i].j2 = ji + 2;
    a[i].j3 = ji + 4;
    ji += 6;
  }
}
int main() {
  int i, j;
  setup();
  while (scanf("%d%d", &n, &k) != EOF) {
    printf("%d\n", a[n].j3 * k);
    for (i = 1; i <= n; i++)
      printf("%d %d %d %d\n", a[i].ou * k, a[i].j1 * k, a[i].j2 * k,
             a[i].j3 * k);
  }
  return 0;
}