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 sum[1100100], s[1100100];
int main() {
  scanf("%d %d %d", &a, &b, &n);
  sum[0] = s[0] = 0;
  long long curr = a;
  for (int i = 1; i < 1100100; i++) {
    sum[i] = sum[i - 1] + curr;
    s[i] = curr;
    curr += b;
  }
  for (int i = 0; i < n; i++) {
    int il, it, im;
    long long l, t, m;
    scanf("%d %d %d", &il, &it, &im);
    l = il;
    t = it;
    m = im;
    int ret =
        min(upper_bound(s, s + 1100100, t) - 1 - s,
            upper_bound(sum, sum + 1100100, t * m + sum[l - 1]) - 1 - sum);
    if (sum[ret] - sum[l - 1] > t * m || s[ret] > t || ret < l)
      printf("-1\n");
    else
      printf("%d\n", ret);
  }
  return 0;
}
