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;
long long A, B, n, l, t, m;
long long h(long long i) { return A + ((i - 1ll) * B); }
bool valid(long long i) {
  if (h(i) > t) return 0;
  long long a = i - l + 1ll;
  long long b = (((i * (i - 1ll)) / 2ll) - ((l - 1ll) * (l - 2ll) / 2ll));
  if (((a * A) + (b * B)) > (m * t)) return 0;
  return 1;
}
long long solve() {
  long long st = l, en = t;
  while (st < en) {
    long long mid = st + (en - st + 1) / 2;
    if (valid(mid)) {
      st = mid;
    } else {
      en = mid - 1;
    }
  }
  return st;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> A >> B >> n;
  for (int i = 0; i < n; i++) {
    cin >> l >> t >> m;
    long long q = solve();
    if (!valid(q))
      cout << -1 << endl;
    else
      cout << q << endl;
  }
  return 0;
}
