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 sigma_size = 26;
const int N = 100 + 50;
const int MAXN = 100000;
const int inf = 0x3fffffff;
const double eps = 1e-8;
const int mod = 1000000000 + 7;
int A, B, n, L, t, m;
int main() {
  while (~scanf("%d%d%d", &A, &B, &n)) {
    for (int i = 1; i <= n; i++) {
      scanf("%d%d%d", &L, &t, &m);
      int l = L, r = inf, ans = -1;
      while (l <= r) {
        int mid = (l + r) >> 1;
        long long sum = (long long)(mid - L + 1) * (long long)A +
                        (long long)(L + mid - 2) * (long long)(mid - L + 1) /
                            2 * (long long)B -
                        (long long)m * (long long)t;
        if ((long long)A + (long long)B * (long long)(mid - 1) <= t &&
            sum <= 0) {
          ans = mid;
          l = mid + 1;
        } else
          r = mid - 1;
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}
