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 Left = 1, n, k, a[10001][4], num, f[10001];
bool b[1000001];
int gf(int i) {
  if (i == f[i]) return i;
  return f[i] = gf(f[i]);
}
int gcd(int a, int b) {
  if (!b) return a;
  return gcd(b, a % b);
}
int main() {
  scanf("%d%d", &n, &k);
  a[1][0] = 1;
  a[1][1] = 2;
  a[1][2] = 3;
  a[1][3] = 5;
  for (int i = 2; i <= n; i++)
    for (int j = 0; j < 4; j++) a[i][j] = a[i - 1][j] + 6;
  printf("%d\n", a[n][3] * k);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= 3; j++) {
      printf("%d", a[i][j] * k);
      if (j != 4) printf(" ");
    }
    printf("\n");
  }
}