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 int A, B, n, t, m, l;
long long tab[1000100];
long long suma[1000100];
void przygotuj() {
  for (int i = 1; i < 1000100; i++) {
    tab[i] = A + B * (i - 1);
    suma[i] = suma[i - 1] + tab[i];
  }
}
int znajdz() {
  int p = l, k = 1000100, mid;
  if (tab[p] > t) return -1;
  while (p + 1 < k) {
    mid = (p + k) / 2;
    if (tab[mid] <= t && suma[mid] - suma[l - 1] <= t * m) {
      p = mid;
    } else
      k = mid;
  }
  return p;
}
int main() {
  cout.sync_with_stdio(0);
  cin >> A >> B >> n;
  przygotuj();
  while (n--) {
    cin >> l >> t >> m;
    int odp = znajdz();
    cout << odp << "\n";
  }
  return 0;
}
