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 months[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
string itos(long long i) {
  string s = "";
  while (i) {
    s += char(i % 10 + '0');
    i /= 10;
  }
  reverse(s.begin(), s.end());
  return s;
}
long long stoi(string &s) {
  long long tot = 0;
  for (int i = (int)s.length() - 1, j = 1; i >= 0; i--, j *= 10) {
    tot += j * (s[i] + '0');
  }
  return tot;
}
bool cmp(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.first > b.first);
}
long long pwr(long long base, long long p, long long mod = 1000000007) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = (ans * base) % 1000000007;
    base = (base * base) % 1000000007;
    p /= 2;
  }
  return ans;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
const int N = 5e5 + 5;
vector<int> v[10000];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(10);
  clock_t clk;
  clk = clock();
  int n, k;
  cin >> n >> k;
  int num = k;
  for (int i = (1); i <= (n); ++i) {
    for (int j = (1); j <= (4); ++j) {
      v[i].push_back(num);
      num += k;
      if (j == 3 || j == 4) num += k;
    }
  }
  cout << v[n][3] << endl;
  for (int i = (1); i <= (n); ++i) {
    for (int j = 0; j < (v[i].size()); ++j) {
      cout << v[i][j] << " ";
    }
    cout << endl;
  }
  clk = clock() - clk;
  cerr << fixed << setprecision(6) << "Time: " << ((double)clk) / CLOCKS_PER_SEC
       << "\n";
  return 0;
}