#include <bits/stdc++.h>
using namespace std;
const int MAXN = -1;
int n, k;
vector<vector<int>> ans;
set<int> S;
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
bool puedo(vector<int> &V, int v) {
  for (int i = (0); i < (int)(V.size()); i++)
    if (gcd(V[i], v) != 1) return false;
  return true;
}
void insertar(int v) {
  for (auto p : S) {
    if (puedo(ans[p], v)) {
      ans[p].push_back(v);
      S.erase(p);
      return;
    }
  }
}
int main() {
  while (scanf("%d %d", &n, &k) == 2) {
    printf("%lld\n", (long long)(6 * n - 1) * k);
    for (int a = (0); a < (int)(n); a++) {
      long long p = 6 * a + 1;
      p *= k;
      long long q = 6 * a + 2;
      q *= k;
      long long r = 6 * a + 3;
      r *= k;
      long long s = 6 * a + 5;
      s *= k;
      printf("%lld %lld %lld %lld\n", p, q, r, s);
    }
  }
  return 0;
}

### Prompt

Your challenge is to write 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>
using namespace std;
const int MAXN = -1;
int n, k;
vector<vector<int>> ans;
set<int> S;
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
bool puedo(vector<int> &V, int v) {
  for (int i = (0); i < (int)(V.size()); i++)
    if (gcd(V[i], v) != 1) return false;
  return true;
}
void insertar(int v) {
  for (auto p : S) {
    if (puedo(ans[p], v)) {
      ans[p].push_back(v);
      S.erase(p);
      return;
    }
  }
}
int main() {
  while (scanf("%d %d", &n, &k) == 2) {
    printf("%lld\n", (long long)(6 * n - 1) * k);
    for (int a = (0); a < (int)(n); a++) {
      long long p = 6 * a + 1;
      p *= k;
      long long q = 6 * a + 2;
      q *= k;
      long long r = 6 * a + 3;
      r *= k;
      long long s = 6 * a + 5;
      s *= k;
      printf("%lld %lld %lld %lld\n", p, q, r, s);
    }
  }
  return 0;
}

```