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;
inline long double min(long double a, long double b) {
  if (a < b) return a;
  return b;
}
inline long double max(long double a, long double b) {
  if (a < b) return b;
  return a;
}
long long A, B;
long long n;
long long sum[1001000];
long long l, t, m;
bool valid(long long pos) {
  long long val = sum[pos] - sum[l - 1];
  if (val <= m * t) return true;
  return false;
}
long long solve() {
  long long high = (t - A) / B + 1;
  long long low = l;
  if (low > high) return -1;
  while (low < high) {
    long long mid = (low + high + 1) / 2;
    if (valid(mid)) {
      low = mid;
    } else
      high = mid - 1;
  }
  if (valid(low)) return low;
  return -1;
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(false);
  cin >> A >> B >> n;
  sum[0] = 0;
  for (int(i) = (1); (i) <= (1e6 + 1); ++(i)) {
    long long val = A + (i - 1) * B;
    sum[i] = sum[i - 1] + val;
  }
  for (int(i) = (0); (i) < (n); ++(i)) {
    cin >> l >> t >> m;
    cout << solve() << endl;
  }
  return 0;
}
