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 long MAX = 1ll << 30;
long long A, B, q;
long long l, t, m;
long long costAt(long long r) {
  long long costR, costL;
  if (r % 2ll == 0)
    costR = (r / 2ll) * (r - 1ll);
  else
    costR = ((r - 1ll) / 2ll) * (r);
  if (l % 2ll == 0)
    costL = ((l - 2ll) / 2ll) * (l - 1ll);
  else
    costL = ((l - 1ll) / 2ll) * (l - 2ll);
  return (r - l + 1ll) * A + (B) * (costR - costL);
}
long long heightAt(long long r) { return A + B * (r - 1ll); }
long long sol() {
  long long left = l - 1, right = MAX;
  while (right - left > 1ll) {
    long long mid = (right + left) >> 1;
    if (heightAt(mid) > t)
      right = mid;
    else
      left = mid;
  }
  if (left == l - 1) return -1;
  right = left + 1, left = l;
  long long lim = m * t;
  while (right - left > 1ll) {
    long long mid = (right + left) >> 1;
    if (costAt(mid) <= lim)
      left = mid;
    else
      right = mid;
  }
  return left;
}
int main() {
  cin >> A >> B >> q;
  while (q--) {
    cin >> l >> t >> m;
    cout << sol() << endl;
  }
}
