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 A, B, n;
int h[1000010];
long long int sum[1000010];
int main() {
  scanf("%d %d %d", &A, &B, &n);
  int maxx = 1;
  while (h[maxx - 1] <= 1e6) {
    h[maxx] = A + (maxx - 1) * B;
    sum[maxx] = sum[maxx - 1] + h[maxx];
    maxx++;
  }
  maxx--;
  for (int i = 1; i <= n; i++) {
    int l;
    long long int t, m;
    scanf("%d %lld %lld", &l, &t, &m);
    int left = l, right = maxx, answer = -1;
    while (left <= right) {
      int mm = (left + right) / 2;
      if (h[mm] <= t && sum[mm] - sum[l - 1] <= 1LL * t * m) {
        answer = mm;
        left = mm + 1;
      } else
        right = mm - 1;
    }
    printf("%d\n", answer);
  }
  return 0;
}
