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;
inline int lowbit(int x) { return x & -x; }
inline int ABS(int x) { return x < 0 ? -x : x; }
inline long long ABS(long long x) { return x < 0 ? -x : x; }
inline void MAX(int &a, int b) {
  if (b > a) a = b;
}
inline void MAX(long long &a, long long b) {
  if (b > a) a = b;
}
inline void MIN(int &a, int b) {
  if (b < a) a = b;
}
inline void MIN(long long &a, long long b) {
  if (b < a) a = b;
}
int fast(int x, int v) {
  int ans = 1;
  for (; v; v >>= 1, x = (long long)x * x % 1000000007)
    if (v & 1) ans = (long long)ans * x % 1000000007;
  return ans;
}
long long fast(long long x, long long v) {
  long long ans = 1;
  for (; v; v >>= 1, x = x * x % 1000000007)
    if (v & 1) ans = ans * x % 1000000007;
  return ans;
}
inline long long sqr(int x) { return x * x; }
int n, k, p, s[10005], top, ans[10005][4], len, t[10005], lt;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    ans[i][0] = p + 1;
    ans[i][1] = p + 2;
    ans[i][2] = p + 3;
    ans[i][3] = p + 5;
    p += 6;
  }
  printf("%I64d\n", (long long)(p - 1) * k);
  for (int i = 1; i <= n; ++i)
    printf("%I64d %I64d %I64d %I64d\n", (long long)ans[i][0] * k,
           (long long)ans[i][1] * k, (long long)ans[i][2] * k,
           (long long)ans[i][3] * k);
}