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 vet[2000010];
bool comp(long long l, long long maior, long long meio, long long m,
          long long t) {
  if (a + ((meio - 1) * b) > t) return false;
  long long an = a + (meio - 1) * b;
  long long sn = ((a + an) * (meio)) / 2;
  long long an1 = a + (l - 2) * b;
  long long sn1 = ((a + an1) * (l - 1)) / 2;
  sn = max(sn, 0LL);
  sn1 = max(sn1, 0LL);
  if (sn - sn1 > m * t) return false;
  return true;
}
int main() {
  cin >> a >> b >> n;
  for (int i = 0; i < 2000008; ++i) {
    if (i == 0) {
      vet[i] = 0;
    } else {
      vet[i] = ((a) + (i - 1) * b) + vet[i - 1];
    }
  }
  for (int i = 0; i < n; ++i) {
    long long l, t, m;
    cin >> l >> t >> m;
    long long menor = l, maior = t + 5, meio;
    while (menor <= maior) {
      meio = (maior + menor) / 2;
      if (comp(l, maior, meio, m, t))
        menor = meio + 1;
      else
        maior = meio - 1;
    }
    if (maior < l)
      cout << "-1" << endl;
    else
      cout << maior << endl;
  }
}
