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>
#pragma comment(linker, "/stack:256000")
using namespace std;
const int N = 100005;
long long getk(int A, int B, int i) { return A + (long long)(i - 1) * B; }
int main() {
  int A, B, q;
  cin >> A >> B >> q;
  while (q--) {
    long long le, lim, m;
    cin >> le >> lim >> m;
    long long k = getk(A, B, le);
    long long maxi = 0, cota = 1500000;
    while (maxi + 1 < cota) {
      long long mid = (maxi + cota) / 2;
      if (2 * mid * k + B * mid * (mid - 1) <= 2 * lim * m) {
        maxi = mid;
      } else {
        cota = mid;
      }
    }
    maxi = min(maxi, (lim - k) / B + 1);
    cout << (maxi >= 1 ? (le + maxi - 1) : -1) << endl;
  }
  return 0;
}
