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;
template <class T>
void debug(T a, T b) {
  ;
}
template <class T>
void chmin(T& a, const T& b) {
  if (a > b) a = b;
}
template <class T>
void chmax(T& a, const T& b) {
  if (a < b) a = b;
}
namespace std {
template <class S, class T>
ostream& operator<<(ostream& out, const pair<S, T>& a) {
  out << '(' << a.first << ',' << a.second << ')';
  return out;
}
}  // namespace std
long long int readL() {
  long long int res;
  scanf("%I64d", &res);
  return res;
}
void printL(long long int res) { printf("%I64d", res); }
long long int A, B, n;
int main() {
  cin >> A >> B >> n;
  for (int hoge = 0; hoge < (n); ++hoge) {
    long long int l = readL(), t = readL(), m = readL();
    --l;
    long long int lb = l, ub = l + 5e6 + 5;
    while (ub - lb > 1) {
      long long int md = (lb + ub) >> 1;
      bool ok = false;
      if (md - l <= m) {
        if (t >= A + B * (md - 1)) {
          ok = true;
        }
      } else {
        long long int sum =
            (A + l * B) * (md - l) + (md - l) * (md - l - 1) / 2 * B;
        if (t >= max(A + B * (md - 1), (sum + m - 1) / m)) {
          ok = true;
        }
      }
      if (ok)
        lb = md;
      else
        ub = md;
    }
    if (lb == l) lb = -1;
    printL(lb);
    putchar('\n');
  }
  return 0;
}
