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 pw(long long b, long long p) {
  if (!p) return 1;
  long long sq = pw(b, p / 2);
  sq *= sq;
  if (p % 2) sq *= b;
  return sq;
}
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
long long sd(long long x) { return x < 10 ? x : x % 10 + sd(x / 10); }
long long lcm(long long a, long long b) { return ((a * b) / gcd(a, b)); }
long long A, B, n, l, t, m, a1, sum;
long long BS() {
  long long b = 0, s = 1, e = 100000000, m;
  while (s <= e) {
    m = s + (e - s) / 2;
    if ((m * (2 * a1 + (m - 1) * B) > 2 * sum) || (a1 + (m - 1) * B > t))
      e = m - 1;
    else {
      b = m;
      s = m + 1;
    }
  }
  if (b)
    return l + b - 1;
  else
    return -1;
}
int main() {
  cin >> A >> B >> n;
  for (int i = 0; i < n; i++) {
    cin >> l >> t >> m;
    a1 = A + (l - 1) * B;
    sum = t * m;
    cout << BS() << endl;
  }
}
