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;
const int MAXN = 100 * 100 * 100 * 3 + 100;
long long a[MAXN], A, B;
int solve(long long x, long long l, long long m, long long t) {
  long long s1, s2;
  s1 = (x - l + 1) * A;
  s2 = ((((x - 1) * (x)) - ((l - 2) * (l - 1))) / 2) * B;
  if (s1 + s2 <= m * t and A + (x - 1) * B <= t)
    return 1;
  else
    return 0;
}
int main() {
  int n;
  cin >> A >> B >> n;
  long long hi, lo, mid, l, t, m;
  vector<int> v;
  for (int i = 0; i < n; i++) {
    cin >> l >> t >> m;
    hi = 10e+9;
    lo = l;
    while (hi - lo > 1) {
      mid = (hi + lo) / 2;
      if (solve(mid, l, m, t))
        lo = mid;
      else
        hi = mid;
    }
    if (solve(hi - 1, l, m, t) == 0 or hi - 1 < l)
      v.push_back(-1);
    else
      v.push_back(hi - 1);
  }
  for (int i = 0; i < v.size(); i++) cout << v[i] << endl;
  return 0;
}
