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, a, b;
bool can(long long l, long long r, long long t, long long m) {
  long long f = a + b * (l - 1), la = a + b * (r - 1);
  long long sum = (f + la) * (r - l + 1) / 2;
  return (la <= t && sum <= m * t);
}
int main() {
  cin >> a >> b >> n;
  while (n--) {
    long long l, le, t, m, r = 1000000;
    r *= r;
    cin >> l >> t >> m;
    le = l;
    while (l <= r) {
      long long mid = (l + r) >> 1;
      if (!can(le, mid, t, m))
        r = mid - 1;
      else
        l = mid + 1;
    }
    long long ans;
    if (r > le)
      ans = r;
    else if (r == le && can(le, r, t, m))
      ans = r;
    else
      ans = -1;
    cout << ans << endl;
  }
  return 0;
}
