Karafs is some kind of vegetable in shape of an 1 × h rectangle. Tavaspolis people love Karafs and they use Karafs in almost any kind of food. Tavas, himself, is crazy about Karafs.

<image>

Each Karafs has a positive integer height. Tavas has an infinite 1-based sequence of Karafses. The height of the i-th Karafs is si = A + (i - 1) × B.

For a given m, let's define an m-bite operation as decreasing the height of at most m distinct not eaten Karafses by 1. Karafs is considered as eaten when its height becomes zero.

Now SaDDas asks you n queries. In each query he gives you numbers l, t and m and you should find the largest number r such that l ≤ r and sequence sl, sl + 1, ..., sr can be eaten by performing m-bite no more than t times or print -1 if there is no such number r.

Input

The first line of input contains three integers A, B and n (1 ≤ A, B ≤ 106, 1 ≤ n ≤ 105).

Next n lines contain information about queries. i-th line contains integers l, t, m (1 ≤ l, t, m ≤ 106) for i-th query.

Output

For each query, print its answer in a single line.

Examples

Input

2 1 4
1 5 3
3 3 10
7 10 2
6 4 8


Output

4
-1
8
-1


Input

1 5 2
1 5 10
2 7 4


Output

1
2
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long A, B, n;
  cin >> A >> B >> n;
  long long l, t, m;
  long long r;
  for (int i = 0; i < n; i++) {
    cin >> l >> t >> m;
    if (A + (l - 1) * B > t)
      cout << -1 << endl;
    else {
      long long p = l;
      r = (t - A) / B + 1;
      while (!((2 * A + B * (r + l - 2)) / 2.0 * (r - l + 1) <= t * m &&
               ((2 * A + B * (r + l)) / 2.0 * (r - l) > t * m))) {
        long long sr = (r + p) / 2;
        if ((2 * A + B * (sr + l - 2)) / 2.0 * (sr - l + 1) <= t * m) {
          p = sr;
        } else
          r = sr;
        if (r - p <= 1) {
          if ((2 * A + B * (r + l - 2)) / 2.0 * (r - l + 1) > t * m) {
            r = p;
          }
          break;
        }
      }
      cout << r << endl;
    }
  }
}
