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;

public class A {

    public static void main(String args[]) {

        Scanner scanner = new Scanner(System.in);
        int a = scanner.nextInt();
        int d = scanner.nextInt();
        int n = scanner.nextInt();

        while (n-- > 0) {
            long l = scanner.nextInt();
            long t = scanner.nextInt();
            long m = scanner.nextInt();


            long start = a + (l - 1) * d;
            long rmin = 1;
            long rmax = t;

            while (rmin < rmax - 1) {
                if (valid(start, d, (rmin + rmax) / 2, m, t)) {
                    rmin = (rmin + rmax) / 2;
                } else {
                    rmax = (rmin + rmax) / 2;
                }
            }

            if (valid(start, d, rmax, m, t)) {
                System.out.println(l - 1 + rmax);
            } else if (valid(start, d, rmin, m, t)) {
                System.out.println(l - 1 + rmin);
            } else System.out.println(-1);
        }

    }

    private static boolean valid(long start, int d, long r, long m, long t) {

        long last = start + (r - 1) * d;
        long sum = (long) r * (start + last) / 2;

//        System.out.println(r + " " + start + " " + last + " " + sum);

        m = r < m ? r : m;

        if (sum > m * t || last > t)
            return false;


        return true;
    }

}
