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>
#pragma comment(linker, "/STACK:100000000000000")
using namespace std;
long long n, k;
vector<long long> ans;
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  scanf("%I64d %I64d", &n, &k);
  long long e = 2, o = 1, mx = 0, tn = n;
  while (tn--) {
    long long a = k * o;
    ans.push_back(k * o);
    o += 2;
    long long b = k * o;
    ans.push_back(k * o);
    o += 2;
    long long c = k * o;
    ans.push_back(k * o);
    mx = max(k * o, mx);
    while (gcd(e * k, a) != k || gcd(e * k, b) != k || gcd(c, k * e) != k)
      e += 2;
    ans.push_back(e * k);
    e += 2;
    mx = max(mx, e - 2);
    o += 2;
  }
  printf("%I64d\n", mx);
  int j = 0;
  while (j < ans.size()) {
    for (int i = 0; i < 4; i++) printf("%I64d ", ans[j++]);
    printf("\n");
  }
  return 0;
}