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;
long long gcd(long long a, long long b) {
  if (b > a) return gcd(b, a);
  if (b == 0) return a;
  return gcd(b, a % b);
}
vector<long long> myv[20000];
long long n, k;
void solve() {
  cin >> n >> k;
  int jj;
  for (int i = 0; i < n; i++) {
    myv[i].push_back(6 * i + 1);
    myv[i].push_back(6 * i + 3);
    myv[i].push_back(6 * i + 5);
    jj = 6 * i + 2;
    myv[i].push_back(jj);
  }
  int ans = k * myv[n - 1][2];
  cout << ans << endl;
  for (int i = 0; i < n; i++) {
    printf("%I64d %I64d %I64d %I64d\n", k * myv[i][0], k * myv[i][1],
           k * myv[i][2], k * myv[i][3]);
  }
}
int main() { solve(); }