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.

<image>

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;
template <typename T, typename U>
ostream& operator<<(ostream& os, const pair<T, U>& p) {
  os << "(" << p.first << "," << p.second << ")";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "[";
  for (__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++i) {
    if (i != v.begin()) os << ", ";
    os << *i;
  }
  os << "]";
  return os;
}
int a, b;
inline long long get_element(int i) { return a + (long long)(i - 1) * b; }
inline long long get_min(int el, int r, int m) {
  long long s = (get_element(r) + get_element(el)) * (r - el + 1) / 2;
  return max((s + m - 1) / m, get_element(r));
}
int main() {
  int n;
  scanf("%d%d%d", &a, &b, &n);
  while (n--) {
    int el, t, m;
    scanf("%d%d%d", &el, &t, &m);
    if (get_min(el, el, m) > t) {
      printf("-1\n");
    } else {
      int u = el, v = t + 1;
      while (v - u > 1) {
        int mid = (u + v) / 2;
        if (get_min(el, mid, m) > t) {
          v = mid;
        } else {
          u = mid;
        }
      }
      printf("%d\n", u);
    }
  }
}
