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;
vector<int> vec[10010];
int ID;
int ord[100];
bool OK;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
void DFS(int dp, int pre, int st) {
  if (OK) return;
  if (dp == 5) {
    OK = 1;
    int i;
    for (i = 1; i <= 4; i++) vec[ID].push_back(ord[i]);
    return;
  }
  int i, j;
  for (i = pre - 1; i >= st; i--) {
    for (j = 1; j < dp; j++)
      if (gcd(ord[j], i) != 1) break;
    if (j == dp) {
      ord[dp] = i;
      DFS(dp + 1, i, st);
    }
  }
}
bool ok(int st, int en) {
  ord[1] = en;
  OK = 0;
  DFS(2, en, st);
  if (OK) return 1;
  return 0;
}
void prepare() {
  int i, now, gas;
  now = 1;
  for (ID = 1; ID <= 10000; ID++) {
    gas = 3;
    while (1) {
      if (ok(now, now + gas)) break;
      gas++;
    }
    now = now + gas + 1;
  }
}
int main() {
  int n, k, i, j, sz;
  prepare();
  scanf("%d %d", &n, &k);
  printf("%d\n", vec[n][0] * k);
  for (i = 1; i <= n; i++) {
    printf("%d", vec[i][0] * k);
    for (j = 1; j < 4; j++) printf(" %d", vec[i][j] * k);
    puts("");
  }
  return 0;
}