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 N = (int)1e6 + 10;
const int MOD = 998244353;
template <class T>
inline bool rd(T &ret) {
  char c;
  int sgn;
  if (c = getchar(), c == EOF) return 0;
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  ret *= sgn;
  return 1;
}
template <class T>
inline void pt(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) pt(x / 10);
  putchar(x % 10 + '0');
}
int main() {
  long long A, B, n, l, t, m;
  rd(A);
  rd(B);
  rd(n);
  while (n-- > 0) {
    rd(l);
    rd(t);
    rd(m);
    long long L = l, R = 1100000;
    while (R - L > 1) {
      int mid = (L + R) >> 1;
      if (A + (mid - 1) * B > t ||
          (mid - l + 1) * (A + (l - 1) * B + A + (mid - 1) * B) / 2 > t * m)
        R = mid;
      else
        L = mid;
    }
    if (A + (L - 1) * B > t ||
        (L - l + 1) * (A + (l - 1) * B + A + (L - 1) * B) / 2 > t * m)
      L = -1;
    pt(L);
    putchar('\n');
  }
  return 0;
}
