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.

<image>

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 double eps = 1e-9;
const double pi = acos(-1.0);
long long a, b;
long long first(long long i) { return a + (i - 1) * b; }
long long get_sum(long long i) {
  if (i == 0) return 0;
  return (a + first(i)) * i / 2LL;
}
long long sum(long long l, long long r) { return get_sum(r) - get_sum(l - 1); }
long long check(long long r, long long m, long long t, long long l) {
  return first(r) <= t && sum(l, r) <= m * t;
}
int main() {
  int n;
  cin >> a >> b >> n;
  while (n--) {
    long long L, t, m;
    cin >> L >> t >> m;
    long long sl = first(L);
    if (sl > t) {
      puts("-1");
      continue;
    }
    long long l = L, r = (1e7), ans = l;
    while (l <= r) {
      long long mid = (l + r) / 2;
      if (check(mid, m, t, L)) {
        l = mid + 1;
        ans = mid;
      } else {
        r = mid - 1;
      }
    }
    cout << ans << endl;
  }
  return 0;
}
