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>
long long bigmod(long long sonkha, long long ghat, long long vag_const) {
  long long vag_shesh = 1;
  while (ghat > 0) {
    if (ghat % 2 == 1) {
      vag_shesh = (vag_shesh * sonkha) % vag_const;
    }
    ghat /= 2;
    sonkha = (sonkha * sonkha) % vag_const;
  }
  return vag_shesh;
}
long long inverse_mod(long long bivajok, long long vag_const) {
  return bigmod(bivajok, vag_const - 2, vag_const);
}
using namespace std;
long long t, m, A, B, l;
bool check(long long r) {
  long long ret = A * (r - l + 1);
  ret += ((((r * (r - 1)) / 2LL) - (((l - 1) * (l - 2)) / 2LL)) * B);
  return ret <= t * min(m, (r - l + 1));
}
int main() {
  long long r, down, up, n, rm, mid;
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> A >> B >> n;
  for (int i = 0; i < n; i++) {
    cin >> l >> t >> m;
    rm = (t - A) / B + 1;
    if (rm < l) {
      cout << -1 << "\n";
      continue;
    }
    r = l;
    up = rm;
    down = l;
    while (down <= up) {
      mid = (up + down) / 2;
      if (check(mid)) {
        r = mid;
        down = mid + 1;
      } else
        up = mid - 1;
    }
    cout << r << "\n";
  }
  return 0;
}
