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;
signed main() {
  ios::sync_with_stdio(0);
  long long A, B, N;
  cin >> A >> B >> N;
  for (long long qi = 0; qi < N; ++qi) {
    long long L, T, M;
    cin >> L >> T >> M;
    long long ans = -1;
    for (long long lb = L, rb = 1LL << 40; lb <= rb;) {
      long long R = lb + rb >> 1;
      long long cost =
          (R - L + 1) * A + (R * (R - 1) / 2 - (L - 1) * (L - 2) / 2) * B;
      if (cost <= min(M, R - L + 1) * T and A + (R - 1) * B <= T) {
        ans = R;
        lb = R + 1;
      } else {
        rb = R - 1;
      }
    }
    cout << ans << endl;
  }
  return 0;
}
