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;
#pragma comment(linker, "/STACK:100000000")
int gcd(int a, int b) {
  if (a)
    return gcd(b % a, a);
  else
    return b;
}
int mas[11111][4];
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) {
    mas[i][0] = i * 6 + 1;
    mas[i][1] = i * 6 + 3;
    mas[i][2] = i * 6 + 5;
  }
  int cnt = 0;
  int idx = 2;
  for (int j = 0; j < n; j++) {
    for (; cnt < n; idx += 2) {
      if (gcd(idx, mas[j][0]) == 1 && gcd(idx, mas[j][1]) == 1 &&
          gcd(idx, mas[j][2]) == 1 && mas[j][3] == 0) {
        mas[j][3] = idx;
        cnt++;
        idx += 2;
        break;
      }
    }
  }
  int maxi = 0;
  for (int j = 0; j < n; j++) {
    maxi = max(maxi, mas[j][0]);
    maxi = max(maxi, mas[j][1]);
    maxi = max(maxi, mas[j][2]);
    maxi = max(maxi, mas[j][3]);
  }
  int m = maxi * k;
  printf("%d\n", m);
  for (int j = 0; j < n; j++) {
    printf("%d %d %d %d\n", mas[j][0] * k, mas[j][1] * k, mas[j][2] * k,
           mas[j][3] * k);
  }
  return 0;
}