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
import java.util.Scanner;
import java.io.OutputStream;
import java.io.IOException;
import java.io.FileOutputStream;
import java.io.PrintWriter;
import java.io.FileInputStream;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }
}

class TaskC {
    public void solve(int testNumber, Scanner in, PrintWriter out) {
        int A = in.nextInt();
        int B = in.nextInt();
        int n = in.nextInt();
        for (int i = 0; i < n; i++) {
            int l = in.nextInt();
            int t = in.nextInt();
            int m = in.nextInt();
            long a = 0;
            long max = 2 * Math.max(m, t) + 1;
            //Bin search
            while (max - a > 1) {
                long q = (max + a) / 2;
                if (A + (l + q - 2) * 1l * B <= t && ((A + (l - 1) * 1l * B) * 1l * q + (q - 1) * 1l * B * 1l * q / 2 <= t * 1l * m)) {
                    a = q;
                } else {
                    max = q;
                }
            }
            if (a == 0) out.println("-1");
            else out.println(l + a - 1);
        }
    }
}
