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.

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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;
const long long INF = 1e18;
const long long mod = 1e9 + 7;
const long long N = 1e5 + 10;
inline long long add(long long x, long long y) {
  x += y;
  if (x >= mod) x -= mod;
  return x;
}
inline long long mul(long long x, long long y) {
  x = (1LL * x * y) % mod;
  return x;
}
long long get_val(long long i, long long a, long long b) {
  return a + b * (i - 1);
}
long long get_sum(long long i, long long a, long long b) {
  return b * (((i + 1) * i) / 2) - b * i + a * i;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long a, b, n;
  cin >> a >> b >> n;
  while (n--) {
    long long l, t, m;
    cin >> l >> t >> m;
    long long l_ = l;
    long long x = get_sum(l - 1, a, b);
    long long r_ = 1e9;
    while (l_ != r_) {
      long long m_ = (l_ - r_) / 2 + r_;
      if (get_val(m_, a, b) <= t && get_sum(m_, a, b) - x <= t * m)
        l_ = m_;
      else
        r_ = m_ - 1;
    }
    if (get_val(l, a, b) > t || get_val(l, a, b) > m * t)
      cout << "-1\n";
    else
      cout << l_ << "\n";
  }
  return 0;
}
