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;
const int INF = 1000000000;
const double PI = 3.1415926535897932384626433832795028841971;
long long sum[10000010], s[10000010];
int main() {
  long long n, m, a, b, l, t, r, j, i;
  cin >> a >> b >> n;
  for (i = 1; i <= 10000000; i++) {
    s[i] = a + (i - 1) * b;
    sum[i] = sum[i - 1] + s[i];
  }
  for (i = 0; i < n; i++) {
    cin >> l >> t >> m;
    int left = l, right = 10000000, middle;
    while (left + 1 != right) {
      middle = (left + right) / 2;
      if (middle < l) {
        left = middle;
        break;
      }
      if (s[middle] <= t && sum[middle] - sum[l - 1] <= t * m)
        left = middle;
      else
        right = middle;
    }
    if (!(s[left] <= t && sum[left] - sum[l - 1] <= t * m)) left = -1;
    if (left < l) left = -1;
    cout << left << endl;
  }
}
