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 = 0x7fffffff;
const int N = 1e5 + 5;
inline long long read() {
  long long x = 0, rev = 0, ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') rev = 1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  return rev ? -x : x;
}
long long a, b, n;
bool judge(long long L, long long R, long long t, long long m) {
  if (a + (R - 1) * b <= t) {
    long long ret = a * (R - L + 1) + (R - L + 1) * (R + L - 2) / 2 * b;
    return ret <= t * m;
  }
  return 0;
}
int main() {
  a = read(), b = read(), n = read();
  while (n--) {
    long long l = read(), t = read(), m = read();
    long long L = l, R = (t - a) / b + 1;
    while (L < R) {
      long long mid = (L + R + 1) >> 1;
      if (judge(l, mid, t, m))
        L = mid;
      else
        R = mid - 1;
    }
    if (a + (L - 1) * b <= t)
      cout << L << endl;
    else
      cout << -1 << endl;
  }
  return 0;
}
