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 int maxt = 1e7 + 10, maxn = 1e5 + 10;
long long a, d;
inline long long A(long long id) { return a + (id - 1LL) * d; }
inline long long second(long long l, long long r) {
  return (r * (A(1) + A(r)) / (2LL) - (l - 1LL) * (A(1) + A(l - 1)) / 2);
}
int main() {
  long long q;
  cin >> a >> d >> q;
  while (q--) {
    long long l, t, m;
    cin >> l >> t >> m;
    if (A(l) > t)
      cout << -1;
    else {
      long long fi = l, se = (1e12);
      while (se - fi > 1) {
        long long mid = (fi + se) / 2;
        if (A(mid) <= t && second(l, mid) <= m * t)
          fi = mid;
        else
          se = mid;
      }
      cout << fi;
    }
    cout << "\n";
  }
  return 0;
}
