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 MAXN = (int)1e6 + 7;
const int INF = (int)1e9 + 7;
long long f, s, n;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> f >> s >> n;
  for (int i = 1; i <= n; i++) {
    long long low, T, m;
    cin >> low >> T >> m;
    long long st = low;
    long long F = f + s * (low - 1);
    long long high = MAXN * 2;
    if (F > T) {
      cout << "-1\n";
      continue;
    }
    while (high - low > 1) {
      long long mid = (low + high) / 2;
      long long S = f + s * (mid - 1);
      long long SUM = (F + S) * (mid - st + 1) / 2;
      if ((SUM + m - 1) / m <= T && S <= T)
        low = mid;
      else
        high = mid;
    }
    cout << low << "\n";
  }
  return 0;
}
