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 warning(disable : 4996)
using namespace std;
int n, m;
set<int> s;
vector<int> a[20000];
int now;
int gcd(int x, int y) { return x == 0 ? y : gcd(y % x, x); }
int main() {
  int i, j, k;
  scanf("%d%d", &n, &m);
  for (i = 0; i < n; i++) {
    a[i].push_back(6 * i + 1);
    a[i].push_back(6 * i + 2);
    a[i].push_back(6 * i + 3);
    a[i].push_back(6 * i + 5);
  }
  printf("%d\n", a[n - 1][3] * m);
  for (i = 0; i < n; i++) {
    for (j = 0; j < 4; j++) printf("%d ", a[i][j] * m);
    printf("\n");
  }
  return 0;
}