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;
long long A, B, l, t, m;
long long X1(int i) { return A + (i - 1) * B; }
bool check(int i) {
  if (X1(i) > t) return false;
  long long S = (X1(l) + X1(i)) * (i - l + 1) / 2;
  long long tmp = max(S % m == 0 ? S / m : S / m + 1, X1(i));
  if (tmp > t) return false;
  return true;
}
int main() {
  int n;
  cin >> A >> B >> n;
  while (n--) {
    cin >> l >> t >> m;
    long long x = A + B * (l - 1);
    if (x > t)
      printf("-1\n");
    else {
      int L = l, R = 1000010;
      while (L < R) {
        int mid = (L + R) >> 1;
        if (L == mid) mid = R;
        if (check(mid))
          L = mid;
        else
          R = mid - 1;
      }
      printf("%d\n", L);
    }
  }
  return 0;
}
