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, l, t, m, sum = 0;
int main() {
  scanf("%lld", &a);
  scanf("%lld", &b);
  scanf("%lld", &n);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &l);
    scanf("%lld", &t);
    scanf("%lld", &m);
    long long end = (t - a) / b + 1;
    if (end < l) {
      cout << "-1\n";
    } else {
      sum = t * m;
      long long ft = a + (l - 1) * b;
      long long sterm = (b - 2 * ft);
      long long sq = (long long)sqrt(sterm * sterm + 8 * b * sum);
      long long r1 = (sterm + sq) / (2 * b);
      long long r2 = (sterm - sq) / (2 * b);
      long long root = max(r1, r2);
      long long ans = l + root - 1;
      printf("%lld\n", min(ans, end));
    }
  }
  return 0;
}
