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;
long long l, t, m;
long long a, b, n;
long long find(long long x) {
  long long lo = 1LL;
  long long hi = 1000000LL;
  while (lo <= hi) {
    long long k = lo + (hi - lo) / 2;
    long long v = k * x + k * (k - 1) * b / 2;
    if (v == m * t) {
      return k;
    }
    if (v < m * t) {
      lo = k + 1;
    }
    if (v > m * t) {
      hi = k - 1;
    }
  }
  return lo - 1;
}
void solve() {
  long long x = 1LL * a + 1LL * (l - 1) * b;
  if (t < x) {
    cout << -1 << endl;
    return;
  }
  long long k = min(find(x), (t - x) / b + 1);
  cout << l + k - 1 << endl;
}
void read() {
  cin >> a >> b >> n;
  for (int i = 0; i < n; ++i) {
    cin >> l >> t >> m;
    solve();
  }
}
int main() {
  read();
  return 0;
}
