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 a, b, n, l, tests, t, m, lo, mid, hi;
long long getSum(long long l, long long r) {
  long long dim = r - l + 1;
  long long ans = dim * (a + b * (l - 1)) + ((dim * (dim - 1)) / 2) * b;
  return ans;
}
long long min(long long a, long long b) { return (a < b ? a : b); }
int main() {
  cin >> a >> b >> tests;
  while (tests--) {
    cin >> l >> t >> m;
    if (getSum(l, l) > t * m) {
      cout << "-1\n";
      continue;
    }
    lo = l - 1;
    hi = 10000LL * 10000LL;
    while (hi - lo > 1) {
      mid = (lo + hi) / 2LL;
      if (getSum(l, mid) <= t * min(m, mid - l + 1) && getSum(mid, mid) <= t)
        lo = mid;
      else
        hi = mid;
    }
    if (lo == l - 1)
      cout << "-1\n";
    else
      cout << lo << "\n";
  }
  return 0;
}
