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 int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
bool has_issue(long long int a, long long int b, long long int c,
               long long int d) {
  if (a < b) {
    if (gcd(a, b) != 1) return true;
  } else {
    if (gcd(b, a) != 1) return true;
  }
  if (a < c) {
    if (gcd(a, c) != 1) return true;
  } else {
    if (gcd(c, a) != 1) return true;
  }
  if (a < d) {
    if (gcd(a, d) != 1) return true;
  } else {
    if (gcd(d, a) != 1) return true;
  }
  return false;
}
int main() {
  long long int n, k, m = -100000;
  cin >> n >> k;
  vector<vector<long long int> > ans(n);
  long long int eve = 2, odd = 1;
  for (int i = 0; i < n; ++i) {
    ans[i].push_back(odd * k);
    odd += 2;
    while (has_issue(eve, odd - 2, odd, odd + 2)) eve += 2;
    ans[i].push_back(eve * k);
    eve += 2;
    ans[i].push_back(odd * k);
    odd += 2;
    ans[i].push_back(odd * k);
    odd += 2;
    if (ans[i][1] > ans[i][3])
      m = ans[i][1];
    else
      m = ans[i][3];
  }
  cout << m << "\n";
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 4; ++j) {
      cout << ans[i][j] << " ";
    }
    cout << "\n";
  }
  return 0;
}