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;
struct rankSet {
  long a, b, c, d;
};
long long gcd(long long a, long long b) {
  if (a > b) {
    long long temp = a;
    a = b;
    b = temp;
  }
  long long c = b % a;
  while (c) {
    b = a;
    a = c;
    c = b % a;
  }
  return a;
}
rankSet getNext(long long firstValidVal) {
  long long a = firstValidVal + 1 - firstValidVal % 2;
  long long b = a + 2;
  long long c = a + 4;
  long long d = firstValidVal + firstValidVal % 2;
  while (gcd(a, d) != 1 || gcd(b, d) != 1 || gcd(c, d) != 1) d += 2;
  rankSet r;
  r.a = a;
  r.b = b;
  r.c = c;
  r.d = d;
  return r;
}
int main() {
  long test_case = 0;
  long long n, k;
  cin >> n >> k;
  rankSet r[10100];
  long long firstvalidVal = 1;
  for (long long i = 0; i < n; i++) {
    r[i] = getNext(firstvalidVal);
    if (r[i].c > r[i].d)
      firstvalidVal = r[i].c + 1;
    else
      firstvalidVal = r[i].d + 1;
  }
  cout << (firstvalidVal - 1) * k << endl;
  for (long long i = 0; i < n; i++) {
    cout << r[i].a * k << " " << r[i].b * k << " " << r[i].c * k << " "
         << r[i].d * k << endl;
  }
  return 0;
}