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 maxn = 100005;
long long A, B;
int n;
int L, t, m;
bool judge(int r) {
  long long s1 = A + (L - 1) * B, s2 = A + (r - 1) * B;
  long long sum = ((r - L + 1) * (s1 + s2)) / 2;
  long long x = max(s2, (sum + m - 1) / m);
  return x <= t;
}
int BS(int l, int r) {
  if (r - l <= 1) {
    if (judge(r)) return r;
    if (judge(l)) return l;
    return -1;
  }
  int m = (l + r) / 2;
  if (judge(m)) return BS(m, r);
  return BS(l, m - 1);
}
int main() {
  cin >> A >> B >> n;
  for (int i = 0; i < n; i++) {
    scanf("%d%d%d", &L, &t, &m);
    printf("%d\n", BS(L, L + 2000000));
  }
  return 0;
}
