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 mod = 1000000007;
int main() {
  long long n, A, B;
  cin >> A >> B >> n;
  int N = 2000011;
  std::vector<long long> h(N);
  std::vector<long long> sumh(N);
  long long curh = A - B;
  h[0] = 0;
  sumh[0] = 0;
  long long curs = 0;
  for (int i = 1; i < N; i++) {
    curh += B;
    h[i] = curh;
    curs += curh;
    sumh[i] = curs;
  }
  long long l, t, m;
  for (int i = 0; i < n; i++) {
    cin >> l >> t >> m;
    long long r1 = upper_bound((h).begin(), (h).end(), t) - h.begin() - 1;
    long long r2 =
        upper_bound((sumh).begin(), (sumh).end(), m * t + sumh[l - 1]) -
        sumh.begin() - 1;
    long long r = min(r1, r2);
    if (r >= l)
      cout << r << endl;
    else
      cout << -1 << endl;
  }
}
