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, l, t, m;
int bin(long long x, long long y) {
  while (x < y) {
    long long mid = (x + y + 1) / 2;
    if (a + (mid - 1) * b <= t &&
        ((mid - l + 1) * a + ((mid * (mid - 1) - (l - 2) * (l - 1)) * b) / 2) <=
            t * m) {
      x = mid;
    } else {
      y = mid - 1;
    }
  }
  return x;
}
void solve() {
  long long n;
  cin >> a >> b >> n;
  for (int i = 0; i < n; i++) {
    cin >> l >> t >> m;
    if (a + (l - 1) * b > t) {
      cout << "-1\n";
      continue;
    }
    cout << bin(l, t) << endl;
  }
}
int main() {
  int z = 1;
  for (int i = 1; i <= z; i++) {
    solve();
  }
}
