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;
long long A, B;
long long solve(long long l, long long t, long long m) {
  long long L, R, mid;
  long long sum = A + B * (l - 1), tot;
  long long temp, h;
  L = 1, R = t - l - A + 2;
  while (L <= R) {
    mid = (L + R) / 2;
    temp = sum * mid + B * mid * (mid - 1) / 2;
    h = sum + (mid - 1) * B;
    if (m > mid)
      tot = t * mid;
    else
      tot = t * m;
    if (h <= t && temp <= tot)
      L = mid + 1;
    else
      R = mid - 1;
  }
  if (L == 1) return -1;
  return l + L - 2;
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  long long n, l, t, m;
  cin >> A >> B >> n;
  for (int i = 0; i < n; i++) {
    cin >> l >> t >> m;
    cout << solve(l, t, m) << '\n';
  }
  return 0;
}
