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 int N = 5e5 + 5;
long long a, b, n, l, t, m;
long long all(long long b) {
  if (b <= 0) return 0;
  return b * (b + 1) / 2;
}
long long as(long long s, long long e) { return (e - s + 1) * a; }
long long bs(long long s, long long e) { return (all(e - 1) - all(s - 2)) * b; }
long long BS() {
  long long s = 0, e = 2e9, mid;
  while (s < e) {
    mid = (s + e + 1) / 2;
    if (as(l, mid) + bs(l, mid) <= m * t && as(mid, mid) + bs(mid, mid) <= t) {
      s = mid;
    } else {
      e = mid - 1;
    }
  }
  return s;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> a >> b >> n;
  while (n--) {
    cin >> l >> t >> m;
    long long ans = BS();
    if (l <= ans)
      cout << ans << "\n";
    else
      cout << "-1\n";
  }
}
