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 mpow(long long a, long long n) {
  long long ret = 1;
  long long b = a;
  while (n) {
    if (n & 1) ret = (ret * b) % 1000000007;
    b = (b * b) % 1000000007;
    n >>= 1;
  }
  return (long long)ret;
}
pair<long long, long long> foo(long long a, long long b, long long l,
                               long long r) {
  return make_pair(((r - l + 1) * (2 * a + (l + r - 2) * b)) / 2ll,
                   a + (r - 1) * b);
}
int main() {
  long long a, b, n;
  cin >> a >> b >> n;
  while (n--) {
    long long l, t, m;
    scanf("%lld", &l), scanf("%lld", &t), scanf("%lld", &m);
    if (a + (l - 1) * b > t) {
      printf("-1\n");
      continue;
    }
    long long mini = l, maxi = (t + 100);
    while (maxi - mini > 1) {
      long long midi = (mini + maxi) / 2;
      pair<long long, long long> p = foo(a, b, l, midi);
      if (p.first > m * t or p.second > t)
        maxi = midi;
      else
        mini = midi;
    }
    printf("%lld\n", mini);
  }
  return 0;
}
