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;
int n, k;
int a[10000][5];
int odd, even;
int v;
int gcd(int a, int b) { return (b ? gcd(b, a % b) : a); }
bool check(int i) {
  int t = even * 2;
  for (int j = 1; j < 4; ++j)
    if (gcd(t, a[i][j]) > 1) return false;
  return true;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; ++i) {
    for (int j = 1; j < 4; ++j) {
      ++odd;
      a[i][j] = odd * 2 - 1;
    }
    do {
      ++even;
      if (check(i)) {
        a[i][0] = even * 2;
        break;
      }
    } while (1);
    for (int j = 0; j < 4; ++j) v = max(v, a[i][j]);
  }
  printf("%d\n", v * k);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < 4; ++j) printf("%d ", a[i][j] * k);
    puts("");
  }
  return 0;
}