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, n;
int solve(long long left, long long t, long long me) {
  long long l = left, r = 1000 * 1000 + 1;
  while (l < r) {
    long long m = (l + r) / 2 + 1;
    if (t >= A + B * (m - 1) &&
        t * me >= ((m - left + 1) * (A + (left - 1) * B) +
                   (m - left) * (m - left + 1) / 2 * B)) {
      l = m;
    } else {
      r = m - 1;
    }
  }
  if (l == r && t < A + B * (l - 1)) {
    return -1;
  } else {
    return r;
  }
}
void solve() {
  cin >> A >> B >> n;
  ;
  for (int i = 0; i < n; i++) {
    long long l, t, m;
    cin >> l >> t >> m;
    cout << solve(l, t, m) << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}
