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 long long inf = 2e18 + 7;
const long long mod = 1e9 + 7;
const double eps = 1e-9;
const double PI = 2 * acos(0.0);
long long a, b, n;
long long l, t, m, ll, q1;
bool f(long long x) {
  if (a + (x - 1) * b <= t) {
    long long sum = a + (x - 1) * b;
    sum = (q1 + sum) * (x - ll + 1) / 2;
    if (sum <= m * t) return true;
  }
  return false;
}
int main() {
  cin >> a >> b >> n;
  for (int i = 0; i < n; i++) {
    cin >> l >> t >> m;
    q1 = a + (l - 1) * b;
    ll = l;
    long long r = 2 * 1e9;
    while (l + 1 < r) {
      long long mid = (l + r) / 2;
      if (!f(mid))
        r = mid;
      else
        l = mid;
    }
    if (f(r)) {
      if (r < ll)
        cout << -1 << endl;
      else
        cout << r << endl;
    } else if (f(l)) {
      if (l < ll)
        cout << -1 << endl;
      else
        cout << l << endl;
    } else
      cout << -1 << endl;
  }
}
