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 long long X = 2e6 + 100;
long long ln;
long long pr[X], lst[X];
long long get_sum(long long l, long long r) {
  if (l == 0) return pr[r];
  return pr[r] - pr[l - 1];
}
long long bin(long long l, long long t, long long m) {
  long long right = ln, left = l;
  while (right - left > 1) {
    long long mid = (left + right) / 2;
    if (get_sum(l, mid) <= t * m && lst[mid] <= t) {
      left = mid;
    } else {
      right = mid;
    }
  }
  if (l >= ln) return -2;
  if (!(get_sum(l, l) <= t * m && lst[l] <= t)) {
    return -2;
  }
  return left;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long a, b, n;
  cin >> a >> b >> n;
  lst[0] = a;
  pr[0] = a;
  ln = X;
  for (long long i = 1; i < X; i++) {
    lst[i] = lst[i - 1] + b;
    pr[i] = pr[i - 1] + lst[i];
    if (lst[i] > X) {
      ln = i;
      break;
    }
  }
  for (long long i = 0; i < n; i++) {
    long long l, t, m;
    cin >> l >> t >> m;
    l -= 1;
    cout << bin(l, t, m) + 1 << '\n';
  }
  return 0;
}
