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;
template <class T>
string to_s(const T& a) {
  ostringstream os;
  os << a;
  return os.str();
}
template <class T>
T to_T(const string& s) {
  istringstream is(s);
  T r;
  is >> r;
  return r;
}
int bin_search(int L, int R, long long A, long long B, long long x) {
  int l = L, r = R, mid = -1;
  long long mid_val;
  while (r >= l) {
    mid = (l + r) / 2;
    mid_val = ((mid - L + 1) * (2 * A + (long long)(mid + L - 2) * B)) / 2;
    if (mid_val < x)
      l = mid + 1;
    else if (mid_val > x)
      r = mid - 1;
    else
      break;
  }
  if (mid_val > x) mid--;
  if (mid >= L && mid <= R)
    return mid;
  else
    return -1;
}
int main() {
  long long A, B, n;
  cin >> A >> B >> n;
  for (int i = (0); i < (long long)(n); i++) {
    long long l, t, m;
    cin >> l >> t >> m;
    int maxr = bin_search(l, l + t + 1, A, B, t * m);
    int max_undert = (t - A) / B + 1;
    if (max_undert < l)
      cout << -1 << endl;
    else
      cout << min(max_undert, maxr) << endl;
  }
  return 0;
}
