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.PrintWriter;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author Artyom Korzun
 */
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++) {
            long l = in.nextInt();
            long t = in.nextInt();
            long m = in.nextInt();

            long answer;
            if (t < computeHeight(a, b, l)) {
                answer = -1;
            } else {
                long maxSum = m * t;
                long left = l;
                long right = (t - a) / b + 2;

                while (left < right - 1) {
                    long mid = (left + right) / 2;
                    long sum = computeSum(a, b, l, mid);
                    if (sum > maxSum)
                        right = mid;
                    else
                        left = mid;
                }

                answer = left;
            }

            out.println(answer);
        }
    }

    private long computeSum(int A, int B, long from, long to) {
        long length = to - from + 1;
        return (computeHeight(A, B, from) + computeHeight(A, B, to)) * length / 2;
    }

    private static long computeHeight(int A, int B, long i) {
        return A + (i - 1) * B;
    }

}

