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 magic = 1000000;
int main() {
  int A, B, n;
  scanf("%d %d", &A, &B);
  scanf("%d", &n);
  int alwaysE = (magic - A) / B + 1;
  for (int i = (0), __ = (n); i < __; ++i) {
    int l, t, m;
    scanf("%d %d", &l, &t);
    scanf("%d", &m);
    int b = l, e = alwaysE;
    int mid, ans = -1;
    while (b <= e) {
      mid = (b + e) / 2;
      int x = (mid - l + 1);
      long long sum = (long long)x * A + (long long)x * l * B +
                      (((-1LL + (x - 2LL)) * x) / 2LL) * B;
      int maior = (mid - 1) * B + A;
      int dv = min(x, m);
      if (max((sum + (dv - 1)) / dv, (long long)maior) <= t) {
        ans = mid;
        b = mid + 1;
      } else
        e = mid - 1;
    }
    printf("%d\n", ans);
  }
  return 0;
}
