#include <bits/stdc++.h>
inline int two(int n) { return 1 << n; }
inline int test(int n, int b) { return (n >> b) & 1; }
inline void set_bit(int& n, int b) { n |= two(b); }
inline void unset_bit(int& n, int b) { n &= ~two(b); }
inline int last_bit(int n) { return n & (-n); }
inline int ones(int n) {
  int res = 0;
  while (n && ++res) n -= n & (-n);
  return res;
}
long long int gcd(long long int a, long long int b) {
  return (a ? gcd(b % a, a) : b);
}
long long int modPow(long long int a, long long int b, long long int MOD) {
  long long int x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y) % MOD;
    }
    b /= 2;
    y = (y * y) % MOD;
  }
  return x;
}
long long int modInverse(long long int a, long long int p) {
  return modPow(a, p - 2, p);
}
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  cout << k * (6 * n - 1) << "\n";
  for (auto i = 0; i < (n); i++)
    cout << k * (6 * i + 1) << " " << k * (6 * i + 2) << " " << k * (6 * i + 3)
         << " " << k * (6 * i + 5) << endl;
}

### Prompt

Generate a Cpp solution to the following 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>.

### Response

```cpp
#include <bits/stdc++.h>
inline int two(int n) { return 1 << n; }
inline int test(int n, int b) { return (n >> b) & 1; }
inline void set_bit(int& n, int b) { n |= two(b); }
inline void unset_bit(int& n, int b) { n &= ~two(b); }
inline int last_bit(int n) { return n & (-n); }
inline int ones(int n) {
  int res = 0;
  while (n && ++res) n -= n & (-n);
  return res;
}
long long int gcd(long long int a, long long int b) {
  return (a ? gcd(b % a, a) : b);
}
long long int modPow(long long int a, long long int b, long long int MOD) {
  long long int x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y) % MOD;
    }
    b /= 2;
    y = (y * y) % MOD;
  }
  return x;
}
long long int modInverse(long long int a, long long int p) {
  return modPow(a, p - 2, p);
}
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  cout << k * (6 * n - 1) << "\n";
  for (auto i = 0; i < (n); i++)
    cout << k * (6 * i + 1) << " " << k * (6 * i + 2) << " " << k * (6 * i + 3)
         << " " << k * (6 * i + 5) << endl;
}

```