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;
int main() {
  long long n, a, b, i, l, t, m, val, s1, s2, low, mid, high, x, y, z, p, q,
      ans;
  while (scanf("%lld%lld%lld", &a, &b, &n) != EOF) {
    for (i = 1; i <= n; i++) {
      scanf("%lld%lld%lld", &l, &t, &m);
      val = t * m;
      x = ((2 * a + (l - 2) * b) * (l - 1)) / 2;
      q = a + (l - 1) * b;
      if (q > t) {
        printf("-1\n");
        continue;
      }
      low = l;
      high = 1000000;
      while (low <= high) {
        mid = (low + high) / 2;
        y = ((2 * a + (mid - 1) * b) * (mid)) / 2;
        p = a + (mid - 1) * b;
        z = y - x;
        if ((z <= val) && (p <= t)) {
          ans = mid;
          low = mid + 1;
        } else
          high = mid - 1;
      }
      printf("%lld\n", ans);
    }
  }
  return 0;
}
