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 sir[5];
int s[10100][5];
int v[10100 * 100], t, o, solt, cnt, n, k;
int gcd(int a, int b) {
  if (!b) return a;
  return gcd(b, a % b);
}
int solve(int k) {
  if (k > 4) {
    o = 1;
    ++solt;
    --n;
    for (register int i = 1; i <= 4; ++i) s[solt][i] = v[sir[i]];
    for (register int i = 1; i <= 4; ++i) v[sir[i]] = 0;
    while (!v[t] && t) --t;
    swap(v[sir[1]], v[t]);
    while (!v[t] && t) --t;
    swap(v[sir[2]], v[t]);
    while (!v[t] && t) --t;
    swap(v[sir[3]], v[t]);
    while (!v[t] && t) --t;
    swap(v[sir[4]], v[t]);
    return o;
  }
  for (register int i = sir[k - 1] + 1; i <= t; ++i) {
    int ok = 1;
    for (register int j = 1; j <= k - 1; ++j)
      if (gcd(v[i], v[sir[j]]) != 1) {
        ok = 0;
        break;
      }
    if (ok) {
      sir[k] = i;
      solve(k + 1);
      if (o) return o;
    }
  }
  return 0;
}
int main() {
  cin >> n >> k;
  for (register int i = 1; i <= 4; ++i) v[i] = i;
  t = 4;
  cnt = 4;
  while (n) {
    o = 0;
    if (solve(1))
      continue;
    else
      v[++t] = ++cnt;
  }
  cout << cnt * k << "\n";
  for (register int i = 1; i <= solt; ++i) {
    for (register int j = 1; j <= 4; ++j) cout << s[i][j] * k << " ";
    cout << "\n";
  }
}