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;
long long L, t, m;
const long long one = 1;
const long long two = 2;
long long get_sum(long long z) {
  if (z < 1) return 0;
  long long res = (z * (z + one)) / two;
  return res;
}
void solve() {
  cin >> L >> t >> m;
  long long num = A + (L - one) * B;
  if (num > t) {
    cout << "-1" << endl;
    return;
  }
  long long LL = L;
  long long RR = 100000000;
  long long R;
  while (true) {
    if (LL == RR) {
      R = LL;
      break;
    }
    long long MM = (LL + RR) / two + one;
    long long number = A + (MM - one) * B;
    if (number <= t)
      LL = MM;
    else
      RR = MM - 1;
  }
  LL = L;
  RR = R;
  long long Max_Number = t * m;
  while (true) {
    if (LL == RR) {
      cout << LL << endl;
      return;
    }
    long long MM = (LL + RR) / two + one;
    long long number = (MM - L + one) * A;
    long long dop = get_sum(MM - one) - get_sum(L - two);
    number += (dop * B);
    if (number <= Max_Number)
      LL = MM;
    else
      RR = MM - 1;
  }
  return;
}
int main() {
  cin >> A >> B >> n;
  for (int i = 1; i <= n; i++) {
    solve();
  }
  return 0;
}
