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 n, m, a, b, c, t, l, times, z1, rem, begin, p, tnew, al, ans, start,
    stop, mid, z;
long long sum(long long no) {
  long long sumis = (no * (2 * al + (no - 1) * b)) / 2;
  return sumis;
}
int main() {
  long long i, j, k;
  cin >> a >> b >> n;
  while (n--) {
    cin >> l >> t >> m;
    al = a + (l - 1) * b;
    start = l;
    z = (t - al) / b + 1;
    stop = l + z - 1;
    if (t < al)
      cout << "-1\n";
    else {
      while (start < stop) {
        mid = (start + stop + 1) / 2;
        if (sum(mid - l + 1) <= m * t) {
          start = mid;
        } else {
          stop = mid - 1;
        }
      }
      cout << start << "\n";
    }
  }
  return 0;
}
