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 initl, t, m;
int main() {
  scanf("%I64d%I64d%I64d", &A, &B, &N);
  while (N--) {
    scanf("%I64d%I64d%I64d", &initl, &t, &m);
    long long start = A + B * (initl - 1);
    if (start > t) {
      puts("-1");
      continue;
    }
    long long l = initl, r = initl + t * m, mid, ans = -1;
    long long maxeat = t * m;
    while (r >= l) {
      mid = (l + r) >> 1;
      long long cur = A + B * (mid - 1);
      long long sum =
          A * (mid - initl + 1) + B * (initl + mid - 2) * (mid - initl + 1) / 2;
      if (cur > t || sum > maxeat)
        r = mid - 1;
      else
        l = mid + 1, ans = mid;
    }
    printf("%I64d\n", ans);
  }
  return 0;
}
