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 double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
long long a, b, n;
long long l, t, m;
long long cal(long long x) { return a + (x - 1) * b; }
long long getsum(long long r) { return (cal(r) + cal(l)) * (r - l + 1) / 2; }
int main() {
  long long i, j, k, maxn;
  while (~scanf("%I64d%I64d%I64d", &a, &b, &n)) {
    while (n--) {
      scanf("%I64d%I64d%I64d", &l, &t, &m);
      if (cal(l) > t) {
        printf("-1\n");
        continue;
      }
      long long ll = l, lr = (t - a) / b + 1, mid;
      while (ll <= lr) {
        long long mid = (ll + lr) / 2;
        if (getsum(mid) <= t * m)
          ll = mid + 1;
        else
          lr = mid - 1;
      }
      printf("%d\n", ll - 1);
    }
  }
  return 0;
}
