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;
#pragma comment(linker, "/STACK:127108864")
const double eps = 1e-9;
const long long inf = 99999999999999999;
const int maxn = 1001;
long long gcd(long long a, long long b) {
  int c;
  if (a < b) swap(a, b);
  while (b != 0) {
    c = a % b;
    a = b;
    b = c;
  }
  return a;
}
int main() {
  long long n, k, cur = 1, cur2 = 2;
  long long a[4][20000];
  cin >> n >> k;
  for (long long i = 0; i < 20000; ++i) {
    a[0][i] = cur;
    cur += 2;
    a[1][i] = cur;
    cur += 2;
    a[2][i] = cur;
    cur += 2;
    while ((gcd(cur2, a[0][i]) != 1) || (gcd(cur2, a[1][i]) != 1) ||
           (gcd(cur2, a[2][i]) != 1))
      cur2 += 2;
    a[3][i] = cur2;
    cur2 += 2;
  }
  cout << k * a[2][n - 1] << '\n';
  for (long long i = 0; i < n; ++i)
    cout << k * a[0][i] << ' ' << k * a[1][i] << ' ' << k * a[2][i] << ' '
         << k * a[3][i] << '\n';
  return 0;
}