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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;
import java.util.Stack;

import static java.lang.Integer.parseInt;

/**
 * TODO: describe
 *
 * @author keks
 */
public class C {
    public static void main(String[] args) throws IOException {
        final BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        final String s1 = reader.readLine();
        String[] split = s1.split("\\s");
        long a = Long.parseLong(split[0]);
        long b = Long.parseLong(split[1]);
        long n = Long.parseLong(split[2]);

        for (int i = 0; i < n; i++) {
            final String s = reader.readLine();
            split = s.split("\\s");
            long ll = Long.parseLong(split[0]);
            long t = Long.parseLong(split[1]);
            long m = Long.parseLong(split[2]);

            long sl = a + (ll - 1) * b;
            long l = ll;
            long r = 2 * t * m;

            if (sl > t) {
                System.out.println(-1);
                continue;
            }

            while (r > l) {
                long mid = l + (r - l) /2;
                if (a + (mid - 1) * b > t) {
                    r = mid;
                    continue;
                }
                long cur = sl * (mid - ll + 1) + (((mid - ll) * (mid - ll + 1)) / 2) * b;

                if (cur <= t * m) {
                    l = mid + 1;
                } else {
                    r = mid;
                }
            }

            System.out.println(l - 1);
        }
    }
}
