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 t, n, i, j, k, x, y, z, l, r, mid, mod = 1e9 + 7, g, m, ans, q;
long long a[100005];
vector<long long> v, u;
string s, st;
bool ok(int mid) {
  long long x = u[mid] - u[l - 1];
  return (x <= m * t);
}
long long bs(long long l, long long r) {
  long long ans = l;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (ok(mid)) {
      ans = mid;
      l = mid + 1;
    } else
      r = mid - 1;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(NULL);
  cin.tie(0);
  cout.tie(0);
  cin >> x >> y >> q;
  u.push_back(0);
  v.push_back(0);
  for (i = 1;; i++) {
    u.push_back(x + y * (i - 1));
    v.push_back(x + y * (i - 1));
    if (x + y * (i - 1) > 1e6) break;
  }
  n = u.size();
  for (i = 1; i < n; i++) u[i] += u[i - 1];
  while (q--) {
    cin >> l >> t >> m;
    r = (upper_bound(v.begin(), v.end(), t) - v.begin()) - 1;
    if (r < l) {
      cout << "-1\n";
      continue;
    }
    ans = bs(l, r);
    cout << ans << "\n";
  }
}
