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;
long long calc(long long l, long long r) {
  return (2 * a + (r + l - 2) * b) * (r - l + 1) / 2;
}
int main() {
  long long n, i, j;
  int l, t, m;
  cin >> a >> b >> n;
  for (i = 0; i < n; i++) {
    scanf("%d%d%d", &l, &t, &m);
    long long A = a + (l - 1) * b;
    long long st = l, ed = 10000000;
    long long sum = 0;
    long long ans = -1;
    while (ed > st + 1) {
      long long mid = (st + ed) / 2;
      sum = calc(l, mid);
      if (max((sum + m - 1) / m, a + (mid - 1) * b) > t)
        ed = mid;
      else
        st = mid;
    }
    sum = calc(l, ed);
    if (max((sum + m - 1) / m, a + (ed - 1) * b) <= t) ans = max(ans, ed);
    sum = calc(l, st);
    if (max((sum + m - 1) / m, a + (st - 1) * b) <= t) ans = max(ans, st);
    cout << ans << endl;
  }
  return 0;
}
