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;
template <class T>
T gcd(T a, T b) {
  T r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
template <class T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
template <class T>
int getbit(T s, int i) {
  return (s >> i) & 1;
}
template <class T>
T onbit(T s, int i) {
  return s | (T(1) << i);
}
template <class T>
T offbit(T s, int i) {
  return s & (~(T(1) << i));
}
const double PI = 2 * acos(0.0);
const double eps = 1e-9;
const int infi = 1e9;
const long long Linfi = (long long)1e18;
const long long MOD = 1000000009;
const int c1 = 31;
int n;
long long A, B, l, t, m;
long long s[2000005], sum[2000005];
int check(int mid) {
  long long minn = min(mid - l + 1, m);
  if (s[mid] <= t && sum[mid] - sum[l - 1] <= minn * t) return 1;
  return 0;
}
void solve() {
  int low = l, high = 2000000, mid, ans = -1;
  while (low <= high) {
    mid = (low + high) >> 1;
    if (check(mid)) {
      low = mid + 1;
      ans = mid;
    } else
      high = mid - 1;
  }
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> A >> B >> n;
  s[1] = A;
  for (int i = 2; i <= 2000005 - 1; i++) s[i] = s[i - 1] + B;
  sum[0] = 0;
  for (int i = 1; i <= 2000005 - 1; i++) sum[i] = sum[i - 1] + s[i];
  while (n--) {
    cin >> l >> t >> m;
    solve();
  }
  return 0;
}
