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 A, B, n;
long long l, t, m;
bool pd__ans(long long l, long long r) {
  long long sl = A + (l - 1) * B;
  long long sr = A + (r - 1) * B;
  if (t >= sr && t * m >= (sl + sr) * (r - l + 1) / 2) return true;
  return false;
}
int main() {
  while (scanf("%I64d%I64d%I64d", &A, &B, &n) != EOF) {
    for (int i = 0; i < n; i++) {
      scanf("%I64d%I64d%I64d", &l, &t, &m);
      long long left = l, right = 100000000LL;
      if (!pd__ans(left, left)) {
        puts("-1");
        continue;
      }
      while (left < right) {
        long long mid = left + (right - left) / 2LL;
        if (pd__ans(l, mid))
          left = mid + 1;
        else
          right = mid;
      }
      printf("%I64d\n", left - 1);
    }
  }
  return 0;
}
