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;
const long long int MOD = 1000000000 + 7;
const long long int MOD1 = 998244353;
long long int pwr(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = res * a;
    a *= a;
    b >>= 1;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
vector<bool> sieve(4000001, true);
void Sieve() {
  sieve[0] = false;
  sieve[1] = false;
  for (long long int i = 2; i * i <= 4000000; i++) {
    if (sieve[i] == true) {
      for (long long int j = i * i; j <= 4000000; j += i) sieve[j] = false;
    }
  }
}
long long int nCr(long long int n, long long int r) {
  long long int ans = 1;
  for (long long int i = (long long int)0; i < (long long int)r; i++) {
    ans = (ans * (n - i));
    ans = (ans / (i + 1));
  }
  return ans;
}
long long int pwrm(long long int a, long long int b, long long int m) {
  a %= m;
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
vector<long long int> sm_pf(10000001);
void sm_pr_fctr() {
  for (long long int i = (long long int)0; i < (long long int)10000001; i++)
    sm_pf[i] = i;
  for (long long int i = 2; i <= 10000000; i++) {
    if (sm_pf[i] == i) {
      for (long long int j = i * i; j <= 10000000; j += i) {
        if (sm_pf[j] == j) sm_pf[j] = i;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int n, k;
  cin >> n >> k;
  long long int ans = (6 * (n - 1) + 5) * k;
  cout << ans << "\n";
  for (long long int i = (long long int)0; i < (long long int)n; i++) {
    cout << (6 * i + 1) * k << " " << (6 * i + 2) * k << " " << (6 * i + 3) * k
         << " " << (6 * i + 5) * k << "\n";
  }
}