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;
struct people {
  int x, y, num;
  bool win;
};
int n, j, i, h, k, q1, q2, q, mx, x[200500], y[200500], cur;
long long a, b, t, m, l, r1, r2, r;
bool check(long long r) {
  long long c = a * (r - l + 1LL) + b * (l - 1LL) * (r - l + 1LL) +
                b * (r - l + 1LL) * (r - l) / 2LL;
  return (a + (r - 1LL) * b <= t && c <= m * t);
}
void solve() {
  scanf("%I64d%I64d%d", &a, &b, &n);
  for (j = 1; j <= n; j++) {
    scanf("%I64d%I64d%I64d", &l, &t, &m);
    if (a + (l - 1) * b > t) {
      printf("-1\n");
      continue;
    }
    r2 = 2000000;
    r1 = l;
    while (r1 < r2) {
      r = (r1 + r2 + 1) >> 1;
      if (check(r)) {
        r1 = r;
      } else {
        r2 = r - 1;
      }
    }
    printf("%d\n", r1);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  srand(333);
  solve();
  return 0;
}
