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 int INF = (-1u) / 2;
const long long int INF2 = (-1ull) / 2;
long long int a, b, i, d[1011000], j, k, n, m, timer = 0, l, r, x, y, t, mid;
long long int c[1011000], cnt = 0, fl = 0, a2, a3 = -1000000, ans = 0;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> a >> b >> n;
  for (i = 0; i < n; i++) {
    cin >> x >> t >> m;
    l = x;
    r = 30000102ll;
    while (l < r) {
      mid = (l + r) / 2ll;
      if (a + (mid - 1ll) * b <= t &&
          ((a + (x - 1ll) * b + a + (mid - 1ll) * b)) * (mid - x + 1) <=
              m * t * 2ll) {
        l = mid + 1;
      } else
        r = mid;
    }
    if (!(a + (r - 1ll) * b <= t &&
          ((a + (x - 1ll) * b + a + (r - 1ll) * b)) * (r - x + 1) <=
              m * t * 2ll))
      r--;
    if (r < x) r = -1;
    cout << r << endl;
  }
  return 0;
}
