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.InputStream;
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.BufferedReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.InputStreamReader;
import java.io.IOException;

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

    static class TaskA {
        public long sum(long l, long r, long A, long B) {
            return A * (r - l + 1) + B * (r * (r - 1) - (l - 1) * (l - 2)) / 2;
        }

        public void solve(int testNumber, InputReader 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 lo = 0;
                long hi = 2l * t * m;
                while (hi - lo > 1) {
                    long mid = (hi + lo) / 2;
                    long r = l + mid;
                    long last = A + (r - 1l) * B;
                    if (last <= t && sum(l, r, A, B) <= m * 1l * t) {
                        lo = mid;
                    } else {
                        hi = mid;
                    }
                }
                long r = l + lo;
                long last = A + (r - 1l) * B;
                if (last <= t && sum(l, r, A, B) <= m * 1l * t) {
                    out.println(l + lo);
                } else {
                    out.println(-1);
                }
            }
        }

    }

    static class InputReader {
        BufferedReader in;
        StringTokenizer tokenizer = null;

        public InputReader(InputStream inputStream) {
            in = new BufferedReader(new InputStreamReader(inputStream));
        }

        public String next() {
            try {
                while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                    tokenizer = new StringTokenizer(in.readLine());
                }
                return tokenizer.nextToken();
            } catch (IOException e) {
                return null;
            }
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

    }
}

