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;
const inline int __GET_INT() {
  int ret;
  scanf("%d", &ret);
  return ret;
}
int main() {
  long long A = __GET_INT(), B = __GET_INT(), n = __GET_INT();
  for (int i = 0; i < n; ++i) {
    long long l = __GET_INT(), t = __GET_INT(), m = __GET_INT();
    long long first = B * (l - 1) + A;
    if (first > t) {
      puts("-1");
    } else {
      int r = -1;
      long long low = 1, high = 1000000, mid;
      while (low <= high) {
        mid = (low + high) >> 1;
        long long val = ((mid * (mid - 1)) / 2) * B + mid * first;
        long long second = B * (l + mid - 2) + A;
        if (second > t || val > (t * m)) {
          high = mid - 1;
        } else {
          low = mid + 1;
          r = l + mid - 1;
        }
      }
      printf("%d\n", r);
    }
  }
  return 0;
}
