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 MAX = 1e8 + 9;
;
long long A, B, n;
long long l, t, m;
long long sum(long long k) { return A * k - B * k + ((k * (k + 1)) / 2) * B; }
bool ok(long long r) {
  long long x = A + B * (r - 1);
  if (x > t) return false;
  long long id = r - m + 1;
  if (id <= l) return true;
  long long rest = t - x;
  rest *= m;
  rest += ((m * (m - 1)) / 2) * B;
  id--;
  long long cur = sum(id) - sum(l - 1);
  if (cur > rest) return false;
  return true;
}
int get() {
  long long from = l;
  long long to = MAX;
  long long mid;
  while (to - from > 1) {
    mid = (from + to) / 2;
    if (ok(mid))
      from = mid;
    else
      to = mid;
  }
  if (ok(to)) return to;
  if (ok(mid)) return mid;
  return from;
}
int main() {
  cin >> A >> B >> n;
  long long r;
  for (int i = 0; i < n; i++) {
    cin >> l >> t >> m;
    if (A + (l - 1) * B > t) {
      cout << "-1\n";
      continue;
    }
    r = get();
    cout << r << endl;
  }
  return 0;
}
