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;
class solution {
 private:
  long long int A, ans, B, n, m, t, l;
  long long int generate_karaf(int i) { return A + (i - 1) * B; }
  long long int sum_of_series(long long int n) {
    return ((n) * ((2 * generate_karaf(l) + (n - 1) * B))) / 2;
  }
  int is_possible(int r) {
    long long int current = generate_karaf(r);
    if (current <= t && sum_of_series(r + 1 - l) <= m * t) {
      return 1;
    }
    return 0;
  }
  void get_maximum_r() {
    if (is_possible(l) == 0) {
      ans = -1;
    }
    int low = l, high = 1000000, i;
    while (low <= high) {
      i = (low + high) / 2;
      if (is_possible(i) == 1) {
        ans = i;
        low = i + 1;
      } else {
        high = i - 1;
      }
    }
  }

 public:
  solution() {
    cin >> A >> B >> n;
    for (int j = 0; j < n; j++) {
      cin >> l >> t >> m;
      get_maximum_r();
      print_answer();
    }
  }
  void print_answer() { cout << ans << endl; }
};
int main() {
  solution *obj = new solution();
  return 0;
}
