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 l, t, m, a, b, n;
long long fun(int l1, int r1) {
  long long n = r1 - l1 + 1;
  long long first = a + (l1 - 1) * b;
  long long sum = (n) * (2 * first + (n - 1) * b) / 2;
  return sum;
}
int main() {
  long long t, index, start, end;
  cin >> a >> b >> n;
  for (int i = 0; i < n; ++i) {
    cin >> l >> t >> m;
    long long x = a + (l - 1) * b;
    if (x > t) {
      printf("-1\n");
    } else {
      index = (t - a) / b + 1;
      if (l == index) {
        printf("%d\n", index);
      } else {
        start = l;
        end = index;
        while (start != end) {
          int mid = (start + end) / 2;
          if (start == mid) break;
          if (fun(l, mid) <= m * t)
            start = mid;
          else
            end = mid;
        }
        if (fun(l, end) <= m * t)
          cout << end << endl;
        else
          cout << start << endl;
      }
    }
  }
  return 0;
}
