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 N = 1e6 + 1;
long long m, n, a, b, sum, l, r, mid, k, t;
bool flag;
long long f(long long y) { return a + b * (y - 1); }
bool judge(long long x) {
  if ((f(k) + f(x)) * (x - k + 1) >> 1 <= t * m) return true;
  return false;
}
int main() {
  scanf("%I64d %I64d %I64d", &a, &b, &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%I64d %I64d %I64d", &k, &t, &m);
    l = k, r = (t - a) / b + 1;
    while (l <= r) {
      mid = l + ((r - l) >> 1);
      if (judge(mid))
        l = mid + 1;
      else
        r = mid - 1;
    }
    if (l - 1 >= k)
      printf("%I64d\n", l - 1);
    else
      printf("-1\n");
  }
}
