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;
const int MAX_S = (int)1e6;
long long a, b;
int max_r;
int n;
int l;
long long t, m;
bool good(int r) {
  if (a + r * b > t) return false;
  long long a0 = a + l * b;
  long long cnt = r - l + 1;
  long long sum = (2 * a0 + b * (cnt - 1)) * cnt / 2;
  return sum <= t * m;
}
void query() {
  scanf("%d%lld%lld", &l, &t, &m);
  l--;
  int left = l - 1, right = max_r + 1;
  while (right - left > 1) {
    int mid = (left + right) / 2;
    if (good(mid))
      left = mid;
    else
      right = mid;
  }
  printf("%d\n", left == l - 1 ? -1 : left + 1);
}
void solve() {
  scanf("%lld%lld%d", &a, &b, &n);
  while (a + (max_r + 1) * b <= MAX_S) max_r++;
  for (int i = 0; i < n; i++) query();
}
int main() {
  solve();
  0;
  return 0;
}
