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>
int nextInt() {
  int x;
  scanf("%d", &x);
  return x;
}
double nextDouble() {
  double x;
  scanf("%lf", &x);
  return x;
}
long long nextLong() {
  long long x;
  scanf("%I64d", &x);
  return x;
}
char nextChar() {
  char x;
  scanf("%c", &x);
  return x;
}
void newLine() { printf("\n"); }
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long pow(long long a, long long b, long long MOD) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y);
      if (x > MOD) x %= MOD;
    }
    y = (y * y);
    if (y > MOD) y %= MOD;
    b /= 2;
  }
  return x;
}
int countSetBit(int n) {
  int ans = 0;
  while (n != 0) {
    n -= (n & -n);
    ++ans;
  }
  return ans;
}
const int N = 10100;
const int M = 10100;
using namespace std;
long long mod = 1ll * (1e9 + 7);
int main() {
  long long n, k;
  cin >> n >> k;
  long long a[N][4];
  long long x = 1;
  for (int i = 0; i < n - 1; i++) {
    a[i][0] = x * k;
    a[i][1] = (x + 1) * k;
    a[i][2] = (x + 2) * k;
    a[i][3] = (x + 4) * k;
    x = x + 6;
  }
  a[n - 1][0] = x * k;
  a[n - 1][1] = (x + 1) * k;
  a[n - 1][2] = (x + 2) * k;
  a[n - 1][3] = (x + 4) * k;
  cout << a[n - 1][3] << endl;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 4; j++) {
      cout << a[i][j] << " ";
    }
    newLine();
  }
}