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;
bool comp(int a, int b) { return (a > b); }
int main() {
  int a, b, n, l, t, m, x, ans, r;
  long long cek;
  scanf("%d %d %d", &a, &b, &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d %d %d", &l, &t, &m);
    if (t < a + (long long)(l - 1) * b)
      printf("-1\n");
    else {
      x = 1;
      ans = l;
      while (x > 0) {
        r = ans + x;
        cek = ((a + (long long)(r - 1) * b + a + (long long)(l - 1) * b) *
               (r - l + 1)) /
              2;
        cek = max(cek / m + (cek % m == 0 ? 0 : 1), a + (long long)(r - 1) * b);
        if (cek <= t)
          ans += x, x *= 2;
        else
          x /= 2;
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}
