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 int Maxn = 10000 + 10;
const int Maxp = 61759 + 100;
int n, k;
vector<long long int> d[Maxn];
bool mark[Maxp];
void input();
void init();
bool gcd(long long int a, long long int b);
void output();
int main() {
  input();
  init();
  output();
  return 0;
}
void input() { cin >> n >> k; }
void init() {
  int a = 0;
  for (int i = 0; i < n; i++) {
    d[i].push_back(6 * a + 1);
    d[i].push_back(6 * a + 2);
    d[i].push_back(6 * a + 3);
    d[i].push_back(6 * a + 5);
    a++;
  }
}
bool gcd(long long int a, long long int b) {
  long long int c;
  while (a != 0) {
    c = a;
    a = b % a;
    b = c;
  }
  return (b == 1);
}
void output() {
  cout << d[n - 1][3] * k << endl;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 4; j++) cout << d[i][j] * k << " ";
    cout << endl;
  }
}