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 INF = INT_MAX;
const long long LINF = LLONG_MAX;
const unsigned long long ULINF = ULLONG_MAX;
const double EPS = 0.000001;
long long solve1(long long a, long long b, long long l, long long t,
                 long long m) {
  double ans = (1. * (4 * a * a - 12 * a * b + 9 * b * b + 8 * a * b * l -
                      12 * b * b * l + 4 * b * b * l * l + 8 * b * m * t)) /
               (b * b);
  ans = pow(ans, 0.5);
  ans = (1. * (ans * b + b - 2 * a)) / (2 * b);
  return floor(ans);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long a, b, n;
  cin >> a >> b >> n;
  while (n--) {
    long long l, t, m;
    cin >> l >> t >> m;
    double right_solution =
        (1. * (-a * a + 3 * a * b - 2 * b * b - 2 * a * b * l + 3 * b * b * l -
               b * b * l * l + b * t + t * t)) /
        (2 * b * t);
    long long ans;
    if (m < right_solution)
      ans = solve1(a, b, l, t, m);
    else
      ans = floor((1. * (-a + b + t)) / b);
    if (ans < l) ans = -1;
    cout << ans << "\n";
  }
  return 0;
}
