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 A, B;
  int N;
  long long l, r, t, tt, M;
  cin >> A >> B >> N;
  for (int i = 0; i < N; i++) {
    scanf("%I64d%I64d%I64d", &l, &t, &M);
    long long st = l, ed = l + 1000000;
    long long H = A + (l - 1) * B;
    long long D = B, C = t / B;
    if (t < H) {
      puts("-1");
      continue;
    }
    while (ed - st > 1) {
      long long md = (st + ed) >> 1;
      long long E = A + (md - 1) * B;
      long long L = md - l + 1;
      long long tot = (H + E) * L / 2;
      long long tt = max(E, (tot - 1 + M) / M);
      if (t >= tt)
        st = md;
      else
        ed = md;
    }
    long long E = A + (ed - 1) * B;
    long long L = ed - l + 1;
    long long tot = (H + E) * L / 2;
    long long tt = max(E, (tot - 1 + M) / M);
    if (t >= tt) st++;
    cout << st << endl;
  }
}
