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 t, n, u, v, m, q, r, ql, qr, k, d, c, s, a, b, l;
const int N = 2e6 + 500;
const long long mod = 1e9 + 7;
const bool JUDGE = false;
bool possible(long long r) {
  if (a + (r - 1LL) * b > t) return false;
  long long sum =
      a * (r - l + 1LL) + b * (((r * (r - 1)) - ((l - 1) * (l - 2))) / 2LL);
  if (sum > m * t) return false;
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  if (JUDGE) {
    freopen("grader.txt", "r", stdin);
    freopen("grader0.txt", "w", stdout);
  }
  cin >> a >> b >> n;
  while (n--) {
    cin >> l >> t >> m;
    long long lo = l, hi = 1e6 + 500;
    while (lo < hi) {
      long long mid = lo + ((hi - lo + 1) >> 1LL);
      if (possible(mid))
        lo = mid;
      else
        hi = mid - 1;
    }
    if (possible(lo))
      cout << lo << '\n';
    else
      cout << -1 << '\n';
  }
  return 0;
}
