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 long double eps = 1e-9;
const int inf = (1 << 30) - 1;
const long long inf64 = ((long long)1 << 62) - 1;
const long double pi = acos(-1);
template <class T>
T sqr(T x) {
  return x * x;
}
template <class T>
T abs(T x) {
  return x < 0 ? -x : x;
}
int main() {
  int aa, bb, n;
  scanf("%d%d%d", &aa, &bb, &n);
  long long a = aa, b = bb;
  for (int i = 0; i < n; ++i) {
    int l, t, m;
    scanf("%d%d%d", &l, &t, &m);
    l--;
    if (a + l * b > t) {
      printf("-1\n");
      continue;
    }
    int ll = l, rr = (t - a) / b;
    while (ll < rr) {
      long long mm = (ll + rr + 1) / 2;
      if ((mm - l + 1) * (a + l * b) + b * (((mm - l + 1) * (mm - l)) / 2) <=
          m * (long long)t) {
        ll = mm;
      } else {
        rr = mm - 1;
      }
    }
    printf("%d\n", rr + 1);
  }
  return 0;
}
