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.

<image>

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;
int main() {
  long long a, b, n;
  cin >> a >> b >> n;
  long long s[1010101] = {0};
  s[0] = 0;
  s[1] = a;
  for (int i = 0; i < 1010101; i++) {
    s[i] = s[i - 1] + a + b * (i - 1);
  }
  while (n--) {
    long long l, t, m;
    cin >> l >> t >> m;
    long long p = t * m;
    long long a1 = l, b1 = 1010100;
    if (t < a + (l - 1) * b) {
      cout << -1 << endl;
      continue;
    }
    while (a1 < b1) {
      long long m1 = (a1 + b1) / 2;
      long long sum = s[m1] - s[l - 1];
      if (a + (m1 - 1) * b > t || sum > p) {
        b1 = m1 - 1;
      } else {
        a1 = m1 + 1;
      }
    }
    long long m1 = (a1 + b1) / 2;
    long long sum = s[m1] - s[l - 1];
    if (a + (m1 - 1) * b <= t && sum <= p) {
      cout << m1 << endl;
    } else
      cout << m1 - 1 << endl;
  }
  return 0;
}
