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 <class _T>
inline string tostr(const _T& a) {
  ostringstream os("");
  os << a;
  return os.str();
}
vector<vector<int> > pre;
int current = 0;
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
bool correct(int x) {
  for (int i = 0; i < pre[current].size(); ++i)
    if (gcd(x, pre[current][i]) != 1) return false;
  return true;
}
int del(int x) {
  int i = 2, num = 0;
  while (i * i <= x) {
    if (x % i == 0) {
      while (x % i == 0) x /= i;
      num++;
    }
    i++;
  }
  if (x != 1) num++;
  return num;
}
int main() {
  pre.resize(1);
  vector<int> tmp;
  tmp.clear();
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i < 1000000; i += 2) {
    if (pre[current].size() == 3) {
      if (current % 2 == 0)
        pre[current].push_back(pre[current][1] / 2 + 1);
      else
        pre[current].push_back(pre[current][1] / 2);
      current++;
      pre.push_back(tmp);
    }
    if (current > n) {
      break;
    }
    if (correct(i)) pre[current].push_back(i);
  }
  int maxi = 0;
  for (int i = 0; i < 3; ++i) maxi = max(maxi, pre[n - 1][i]);
  printf("%d\n", maxi * k);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 4; ++j) printf("%d ", pre[i][j] * k);
    printf("\n");
  }
  return 0;
}