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 = 1e5 + 5;
const int mod = 1e9 + 7;
long long n, A, B, m, a, t;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> A >> B >> n;
  for (int i = 0; i < n; i++) {
    cin >> a >> t >> m;
    long long l = a, r = 10000000000000LL + 1 + a;
    while (l + 1 < r) {
      long long mid = (l + r) / 2;
      long long sum =
          (((mid - 2 + a) * (mid + 1 - a) / 2) * B) + A * (mid - a + 1);
      if (sum <= t * m && A + B * (mid - 1) <= t)
        l = mid;
      else
        r = mid;
    }
    if (l != a)
      cout << l << '\n';
    else if (A + B * (l - 1) <= t)
      cout << l << '\n';
    else
      cout << -1 << '\n';
  }
  return 0;
}
