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;
int n, a, b;
long long height(long long i) { return (1ll) * a + (i - 1) * b; }
long long sum(long long l, long long r) {
  return a * (1ll) * (r - l + 1) +
         b * (r * (r - 1) / 2 - ((l - 1) * (l - 2) / 2));
}
int main() {
  cin >> a >> b >> n;
  for (long long i = 0; i < n; i++) {
    int l, t, m;
    cin >> l >> t >> m;
    if (height(l) > t)
      printf("%d\n", -1);
    else {
      long long left = l, right = 2e6 + 5;
      while (right - left > 1) {
        long long mid = (left + right) / 2;
        if (height(mid - 1) <= t && sum(l, mid - 1) <= (1ll) * t * m)
          left = mid;
        else
          right = mid;
      }
      cout << left - 1 << "\n";
    }
  }
}
