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;
long long iter, wide, lllft, lef;
bool isOk(long long rig) {
  long long total =
      (A + B * (lllft - 1) + A + B * (rig - 1)) * (rig + 1 - lllft) / 2;
  long long maxx = A + B * (rig - 1);
  return total <= iter * wide && maxx <= iter;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  int n;
  while (cin >> A) {
    cin >> B >> n;
    while (n--) {
      long long rig, step;
      cin >> lllft >> iter >> wide;
      rig = lllft - 1;
      step = iter;
      while (step) {
        if (isOk(rig + step)) {
          rig += step;
        } else
          step /= 2;
      }
      if (rig < lllft) rig = -1;
      cout << rig << "\n";
    }
  }
}
