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 N = 3e6 + 5;
long long int V[N];
long long int A, B, n;
long long int check(long long int r, long long int l, long long int t,
                    long long int m) {
  long long int mx = (A + (r - 1LL) * B);
  long long int s = V[r] - V[l - 1];
  return (s <= (m * t) and mx <= t);
}
long long int Q(long long int l, long long int t, long long int m) {
  long long int lo = l, hi = N, ans = -1;
  while (lo <= hi) {
    long long int mid = (lo + hi) >> 1;
    if (check(mid, l, t, m)) {
      ans = mid;
      lo = mid + 1;
    } else {
      hi = mid - 1;
    }
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> A >> B >> n;
  for (int i = 1; i <= 3e6; i++) {
    V[i] = A + (i - 1LL) * B;
    V[i] += V[i - 1];
  }
  for (int i = 0; i < n; i++) {
    long long int l, t, m;
    cin >> l >> t >> m;
    cout << Q(l, t, m) << '\n';
  }
  return 0;
}
