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;
template <typename Head>
void print(Head&& head) {
  cout << head << '\n';
}
template <typename Head, typename... Tail>
void print(Head&& head, Tail... tail) {
  cout << head << " ";
  print(tail...);
}
long long int mceil(long long int a, long long int b) {
  return (a / b + (a % b != 0));
}
long long int mfloor(long long int a, long long int b) { return (a / b); }
long long modmul(long long a, long long b, long long MOD = 1000000007) {
  return ((a % MOD) * (b % MOD)) % MOD;
}
long long modadd(long long a, long long b, long long MOD = 1000000007) {
  return ((a % MOD) + (b % MOD) + MOD) % MOD;
}
long long modsub(long long a, long long b, long long MOD = 1000000007) {
  return ((a % MOD) - (b % MOD) + MOD) % MOD;
}
long long int fastexpo(long long int a, long long int b,
                       long long int MOD = 1000000007) {
  a = a % MOD;
  long long int ans = 1;
  while (b) {
    if (b & 1) ans = (ans * 1ll * a) % MOD;
    a = (a * 1ll * a) % MOD;
    b = b / 2;
  }
  return ans;
}
vector<int> all_div[200001];
void alldivisors() {
  for (int i = 1; i <= 200000; ++i) {
    for (int j = i; j <= 200000; j += i) all_div[j].push_back(i);
  }
}
bool ispower2(long long int n) { return (n and (n & (n - 1)) == 0); }
long long int summ(long long int n) {
  long long int sum = 0;
  while (n) {
    sum += n % 10;
    n /= 10;
  }
  return sum;
}
long long int myfloor(long long int n, long long int k) {
  long long int ans = n / k;
  if (n % k and n < 0) ans--;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int tt = 1;
  for (long long int ii = 0; ii < tt; ii++) {
    long long int n, k;
    cin >> n >> k;
    print((6 * n - 1) * k);
    for (long long int i = 0; i < n; i++) {
      print(k * (6 * i + 1), k * (6 * i + 2), k * (6 * i + 3), k * (6 * i + 5));
    }
  }
  return 0;
}