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 N = 2000500;
const int INF = 1000000000, mod = 1000000007;
const long long LLINF = 1000000000000000000ll;
long long seq[N];
long long sum[N];
int main() {
  long long a, b;
  cin >> a >> b;
  seq[1] = a;
  for (int i = 2; i < N; i++) seq[i] = seq[i - 1] + b;
  sum[1] = seq[1];
  for (int i = 2; i < N; i++) sum[i] = sum[i - 1] + seq[i];
  int q;
  cin >> q;
  for (; q != 0; --q) {
    long long l, m, t;
    cin >> l >> t >> m;
    int ll = 0, rr = N - 1;
    while (rr - ll > 1) {
      int mm = (ll + rr) / 2;
      if (sum[mm] - sum[l - 1] <= m * t && seq[mm] <= t)
        ll = mm;
      else
        rr = mm;
    }
    cout << (ll < l ? -1 : ll) << '\n';
  }
  return 0;
}
