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 int part_sum(long long int a, long long int b, long long int l,
                       long long int r) {
  return (2 * (a + (l - 1) * b) + (r - l) * b) * (r - l + 1) / 2;
}
int main() {
  long long int a, b, n;
  cin >> a >> b >> n;
  for (auto i = 0; i < n; ++i) {
    long long int l, t, m;
    cin >> l >> t >> m;
    if (a + (l - 1) * b > t) {
      cout << "-1\n";
    } else {
      long long int max_r = 1 + (t - a) / b;
      long long int left = l, right = max_r;
      while (right - left > 1) {
        long long int mid = (left + right) / 2;
        long long int psm = part_sum(a, b, l, mid);
        if (psm > t * m) {
          right = mid;
        } else {
          left = mid;
        }
      }
      if (part_sum(a, b, l, right) <= t * m) {
        cout << right << endl;
      } else {
        cout << right - 1 << endl;
      }
    }
  }
  return 0;
}
