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 int A, B;
int N;
long long int getIthSum(long long int r) {
  return r * A + (((r - 1) * r) / 2) * B;
}
bool satisfies(long long int l, long long int r, long long int t,
               long long int m) {
  long long int rElement = (r - 1) * B + A;
  if (rElement > t) {
    return false;
  }
  long long int lElement = (l - 1) * B + A;
  long long int sum = getIthSum(r) - getIthSum(l) + lElement;
  if (sum > m * t) {
    return false;
  }
  return true;
}
int smallestR(long long int l, long long int t, long long int m) {
  long long int leftmost = l;
  long long int rightmost = 1000000;
  long long int mid = (leftmost + rightmost) / 2;
  while (rightmost - leftmost > 2) {
    if (satisfies(l, mid, t, m)) {
      leftmost = mid;
    } else {
      rightmost = mid;
    }
    mid = (leftmost + rightmost) / 2;
  }
  int maxRes = -1;
  for (int i = leftmost; i <= rightmost; i++) {
    if (satisfies(l, i, t, m)) {
      maxRes = i;
    }
  }
  return maxRes;
}
int main() {
  cin >> A >> B >> N;
  for (int i = 0; i < N; i++) {
    long long int l, t, m;
    cin >> l >> t >> m;
    cout << smallestR(l, t, m) << endl;
  }
  return 0;
}
