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 = 1000000007;
const long long int inf = 1e17;
const long long int N = 1e5 + 5;
long long int mult(long long int a, long long int b) {
  a %= mod;
  b %= mod;
  long long int res = a;
  res *= b;
  if (res >= mod) {
    res %= mod;
  }
  return res;
}
long long int expo(long long int base, long long int exponent) {
  long long int ans = 1;
  base %= mod;
  while (exponent != 0) {
    if (exponent & 1) ans = (1LL * ans * base) % mod;
    base = (1LL * base * base) % mod;
    exponent >>= 1;
  }
  return ans % mod;
}
long long int inv_mod(long long int k) { return expo(k, mod - 2); }
long long int calc(long long int x) {
  x %= mod;
  return (mult(x, mult(x + 1, inv_mod(2LL))));
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n, k;
  cin >> n >> k;
  long long int ans[n][4];
  memset(ans, 0, sizeof(ans));
  for (long long int i = 0; i < n; i++) {
    for (long long int j = 0; j < 4; j++) {
      ans[i][j] = 6 * i + (2 * j + 1);
    }
  }
  for (long long int i = 0; i < n; i++) {
    ans[i][3] = 6 * i + 2;
  }
  long long int res = 0;
  for (long long int i = 0; i < n; i++) {
    for (long long int j = 0; j < 4; j++) {
      res = max(res, k * ans[i][j]);
    }
  }
  cout << res << '\n';
  for (long long int i = 0; i < n; i++) {
    for (long long int j = 0; j < 4; j++) {
      cout << ans[i][j] * k << ' ';
    }
    cout << '\n';
  }
}