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 n;
long long a, b, t, m;
long long sum[1000100];
void init() {
  sum[0] = 0;
  sum[1] = a;
  for (long long i = 2; i <= 1000000; i = i + 1LL)
    sum[i] = sum[i - 1] + a + (i - 1LL) * b;
}
int main() {
  long long tmp;
  scanf("%I64d%I64d%I64d", &a, &b, &n);
  init();
  while (n--) {
    scanf("%I64d%I64d%I64d", &tmp, &t, &m);
    long long left = tmp, right = 1000000;
    while (left < right) {
      long long mid = left + (right - left + 1LL) / 2LL;
      if ((a + (mid - 1LL) * b) <= t && (sum[mid] - sum[tmp - 1]) <= t * m)
        left = mid;
      else
        right = mid - 1LL;
    }
    if ((a + (left - 1LL) * b) <= t && (sum[left] - sum[tmp - 1]) <= t * m)
      printf("%I64d\n", left);
    else
      puts("-1");
  }
  return 0;
}
