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 a, b, n, x, y, z;
int main() {
  scanf("%I64d %I64d %I64d", &a, &b, &n);
  for (int i = 0; i < n; i++) {
    scanf("%I64d %I64d %I64d", &x, &y, &z);
    long long st = x;
    long long ed = 10000000LL;
    long long zz = a * (st - 1) + b * ((st - 1) * (st - 2) / 2);
    while (st <= ed) {
      long long mid = (st + ed) / 2;
      long long xx = a + b * (mid - 1);
      long long yy = a * mid + b * (mid * (mid - 1) / 2) - zz;
      if (xx <= y && yy <= y * z) {
        st = mid + 1;
      } else {
        ed = mid - 1;
      }
    }
    --st;
    if (st < x) {
      st = -1;
    }
    printf("%I64d\n", st);
  }
  return 0;
}
