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 double EPS = 1e-7;
const int MOD = (int)1e9 + 7;
const int MAXN = (int)1e5 + 10;
long long A, B, n;
long long solve(long long l, long long m, long long k) {
  long long ans = A + (l + k - 1) * B;
  long long val = (k + 1) * (A + (l - 1) * B) + k * (k + 1) * B / 2;
  ans = max(ans, (val + m - 1) / m);
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> A >> B >> n;
  for (int i = 0; i < n; ++i) {
    long long l, t, m;
    cin >> l >> t >> m;
    if (solve(l, m, 0) > t) {
      cout << -1 << endl;
      continue;
    }
    long long lo = 0, hi = 1;
    while (solve(l, m, hi) <= t) hi *= 2;
    while (lo + 1 < hi) {
      int mid = (lo + hi) >> 1;
      if (solve(l, m, mid) <= t)
        lo = mid;
      else
        hi = mid;
    }
    cout << l + lo << endl;
  }
  return 0;
}
