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;
int main() {
  int a, b, n;
  scanf("%d%d%d", &a, &b, &n);
  for (int i = 1; i <= n; i++) {
    long long l, t, m;
    cin >> l >> t >> m;
    long long sum1;
    if (l == 1)
      sum1 = 0;
    else
      sum1 = (l - 1) * (2 * a + (l - 2) * b) / 2;
    long long le = l, ri = 100000000;
    long long mid;
    long long flag = -1;
    while (le < ri) {
      mid = (le + ri) / 2;
      if (a + b * (mid - 1) <= t &&
          (mid) * (a * 2 + b * (mid - 1)) / 2 - sum1 <= t * m) {
        le = mid + 1;
        flag = max(flag, mid);
      } else
        ri = mid;
    }
    cout << flag << endl;
  }
  return 0;
}
