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.

<image>

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
import java.util.Scanner;

public class Solution {
    public static void main(String[] args) {
        Scanner reader = new Scanner(System.in);
        long A = reader.nextInt();
        long B = reader.nextInt();
        int n = reader.nextInt();
        for (int i = 0; i < n; i++) {
            int l = reader.nextInt();
            int t = reader.nextInt();
            int m = reader.nextInt();
            long a = 0, b = 2 * Math.max(t, m) + 1;
            while (b - a > 1) {
                long q = (a + b) / 2;
                if (A + (l + q - 2) * B <= t && (A + (l - 1) * B) * q + B * q * (q - 1) / 2 <= (long) m * t)
                    a = q;
                else
                    b = q;
            }
            System.out.println((a == 0) ? -1 : l + a - 1);
        }
    }
}
