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;
int main() {
  long long a, b, n;
  cin >> a >> b >> n;
  vector<long long> v(1000001);
  v[1] = a;
  for (int i = 2; i <= 1000000; i++) v[i] = v[i - 1] + b;
  vector<long long> w(1000001);
  w[0] = 0;
  for (int i = 1; i < 1000000; i++) w[i] = w[i - 1] + v[i];
  int q1, q2, q3;
  long long l, t, m, f, s;
  while (n > 0) {
    n--;
    scanf("%d%d%d", &q1, &q2, &q3);
    l = q1;
    t = q2;
    m = q3;
    if (v[l] > t)
      cout << "-1" << endl;
    else {
      f = t * m + w[l - 1];
      s = upper_bound(w.begin(), w.end(), f) - w.begin();
      f = upper_bound(v.begin(), v.end(), t) - v.begin();
      if (f < s) s = f;
      cout << s - 1 << endl;
    }
  }
  return 0;
}
