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 double EPS = 1e-9;
long long int solve(long long int a, long long int b, long long int t,
                    long long int m) {
  if (a > t) {
    return -1;
  }
  long long int lo = 0;
  long long int hi = (t - a) / b + 2;
  while (hi - lo >= 2) {
    long long int mid = (lo + hi) / 2;
    bool ok = mid * (mid - 1) / 2 * b + a * mid <= m * t;
    if (mid <= m) {
      ok &= a + (mid - 1) * b <= t;
    }
    if (ok) {
      lo = mid;
    } else {
      hi = mid;
    }
  }
  return lo;
}
int main(void) {
  int a, b, n;
  cin >> a >> b >> n;
  for (int loop_var = (int)(0); loop_var < (int)(n); loop_var++) {
    long long int l, t, m;
    cin >> l >> t >> m;
    int res = solve((l - 1) * b + a, b, t, m);
    cout << (res == -1 ? -1 : res + l - 1) << endl;
  }
}
