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;
long long need[2000001];
bool possible(long long fidx, long long tidx, long long first, long long change,
              long long limit, long long dist) {
  bool ans = false;
  long long startval = first + (fidx - 1) * change;
  long long endval = first + (tidx - 1) * change;
  if (endval > limit) {
    return false;
  }
  long long total = ((tidx - fidx + 1) * (startval + endval)) / 2;
  long long xx = limit * dist;
  if (total <= xx) {
    return true;
  }
  return false;
}
int main(void) {
  int t = 1;
  while (t--) {
    long long A, B, n;
    cin >> A >> B >> n;
    for (int i = 1; i <= n; i++) {
      long long l, t, m;
      scanf("%I64d", &l);
      scanf("%I64d", &t);
      scanf("%I64d", &m);
      long long low = l;
      long long high = l + t;
      long long ans = -1;
      while (low <= high) {
        long long mid = (low + high) / 2;
        if (possible(l, mid, A, B, t, m)) {
          ans = max(ans, mid);
          low = mid + 1;
        } else {
          high = mid - 1;
        }
      }
      printf("%I64d\n", ans);
    }
  }
  return 0;
}
