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 inf = (int)1e9;
const int mod = inf + 7;
const double eps = 1e-9;
const double pi = acos(-1.0);
int x, d, n;
long long get(int ind) { return x + (ind - 1) * 1ll * d; }
long long sum(int l, int r) {
  return (r - l + 1) * 1ll * (get(l) + get(r)) / 2;
}
bool ok(int l, int r, int t, int m) {
  return get(r) <= t && sum(l, r) <= t * 1ll * m;
}
void solve() {
  int l, t, m;
  scanf("%d%d%d", &l, &t, &m);
  int L = l, R = (int)1e6, ans = -1;
  while (L <= R) {
    int mid = (L + R) >> 1;
    if (ok(l, mid, t, m)) {
      ans = mid;
      L = mid + 1;
    } else
      R = mid - 1;
  }
  printf("%d\n", ans);
}
int main() {
  scanf("%d%d%d", &x, &d, &n);
  while (n--) solve();
  return 0;
}
