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;
vector<long long> sum, arr;
vector<long long>::iterator it;
int main() {
  int n, i, a, b, l, r1, r2;
  long long x, k, y, t, m, val;
  scanf("%d%d%d", &a, &b, &n);
  arr.clear();
  sum.clear();
  x = 0LL;
  y = 0LL;
  arr.push_back(y);
  sum.push_back(x);
  x = y = (long long)a;
  arr.push_back(y);
  sum.push_back(x);
  for (i = 1; i < 1000006; i++) {
    y = y + (long long)b;
    x = x + y;
    arr.push_back(y);
    sum.push_back(x);
  }
  while (n--) {
    scanf("%d%I64d%I64d", &l, &t, &m);
    if (arr[l] > t) {
      printf("-1\n");
      continue;
    }
    long long prod = t * m;
    val = sum[l - 1] + prod;
    it = lower_bound(sum.begin(), sum.end(), val);
    r1 = it - sum.begin();
    if (*it != val) r1--;
    val = t;
    it = lower_bound(arr.begin(), arr.end(), val);
    r2 = it - arr.begin();
    if (*it != val) r2--;
    r1 = min(r1, r2);
    printf("%d\n", r1);
  }
  return 0;
}
