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>
const double EPS = 1e-24;
const long long MOD = 1000000007ll;
const double PI = 3.14159265359;
int INF = 2147483645;
template <class T>
T Max2(T a, T b) {
  return a < b ? b : a;
}
template <class T>
T Min2(T a, T b) {
  return a < b ? a : b;
}
template <class T>
T Max3(T a, T b, T c) {
  return Max2(Max2(a, b), c);
}
template <class T>
T Min3(T a, T b, T c) {
  return Min2(Min2(a, b), c);
}
template <class T>
T Max4(T a, T b, T c, T d) {
  return Max2(Max2(a, b), Max2(c, d));
}
template <class T>
T Min4(T a, T b, T c, T d) {
  return Min2(Min2(a, b), Max2(c, d));
}
using namespace std;
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  long long a, b, n, i, j, k, l;
  cin >> a >> b >> n;
  for (i = 0; i < n; i++) {
    long long l, t, m;
    cin >> l >> t >> m;
    long long s = l, e = 1e12;
    long long sl = a + (l - 1) * b;
    while (s + 1 < e) {
      long long r = (s + e) / 2;
      long long sr = a + (r - 1) * b;
      long long sum = a * (r - l + 1);
      sum += b * (((r - l + 1) * (l + r - 2)) / 2);
      if (sr <= t && sum <= (m * t))
        s = r;
      else
        e = r;
    }
    bool fail = false;
    if (s == l) {
      if (sl > t) {
        cout << "-1\n";
        fail = true;
      }
    }
    if (!fail) cout << s << endl;
  }
  return 0;
}
