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 A, B, N;
long long calc(int a) {
  return (long long)A + (long long)(a - 1) * (long long)B;
}
int main() {
  scanf("%d%d%d", &A, &B, &N);
  for (int i = 1; i <= N; i++) {
    int L, t, m;
    scanf("%d%d%d", &L, &t, &m);
    long long cap = (long long)t * (long long)m;
    int l = L, r = l + 1000000;
    while (l < r) {
      if (l + 1 == r) break;
      int mid = (l + r) >> 1;
      if (calc(mid) <= t &&
          (calc(L) + calc(mid)) * (long long)(mid - L + 1) / 2LL <= cap)
        l = mid;
      else
        r = mid;
    }
    if (calc(l) <= t &&
        (calc(L) + calc(l)) * (long long)(l - L + 1) / 2LL <= cap)
      printf("%d\n", l);
    else
      puts("-1");
  }
  getchar();
  getchar();
  return 0;
}
