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;
int main() {
  long long a, b, n;
  cin >> a >> b >> n;
  for (int ii = 0; ii < n; ii++) {
    int l, t, m;
    cin >> l >> t >> m;
    long long L = a + (l - 1) * b;
    if (L > t) {
      cout << -1 << endl;
      continue;
    }
    long long target = (long long)t * m;
    long long left = 0, right = (t - L) / b;
    while (left < right) {
      long long mid = left + (right - left) / 2;
      long long R = L + mid * b;
      long long current = (L + R) * (mid + 1) / 2;
      if (current <= target && current + R + b > target) {
        left = right = mid;
        break;
      }
      if (current < target)
        left = mid + 1;
      else
        right = mid - 1;
    }
    cout << l + left << endl;
  }
  return 0;
}
