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;
using namespace std;
long long a, b, n, l, t, m, L, R, M, i;
int main() {
  cin >> a >> b >> n;
  while (n--) {
    cin >> l >> t >> m;
    if (t < (a + (l - 1) * b)) {
      cout << -1 << endl;
      continue;
    }
    L = 1;
    R = 10000000;
    while (L <= R) {
      M = (L + R) >> 1;
      if ((M - l + 1) * (2 * a + (M + l - 2) * b) <= 2 * m * t)
        L = M + 1;
      else
        R = M - 1;
    }
    cout << min(R, (t - a + b) / b) << endl;
  }
}
