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 sum(long long a, long long b, long long r) {
  if (r == 0) return 0;
  long long ret = a * r + b * (r - 1) * r / 2;
  return ret;
}
inline long long h(long long i, long long a, long long b) {
  return a + (i - 1LL) * b;
}
int main() {
  long long a, b, n;
  cin >> a >> b >> n;
  for (int i = 0; i < n; i++) {
    long long l, t, m;
    cin >> l >> t >> m;
    if (h(l, a, b) > t) {
      cout << -1 << endl;
      continue;
    }
    long long x = 1, y = 10000000;
    long long indx = -1;
    while (x <= y) {
      long long mid = (x + y) / 2ll;
      long long s = sum(a, b, mid) - sum(a, b, l - 1);
      if (s <= t * m && h(mid, a, b) <= t) {
        x = mid + 1;
        indx = mid;
      } else {
        y = mid - 1;
      }
    }
    cout << indx << endl;
  }
}
