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

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);
        TaskС solver = new TaskС();
        solver.solve(1, in, out);
        out.close();
    }
}

class TaskС {
    private long a, b, n;

    long S(long i) {
        return a + b*(i-1L);
    }

    long sum(long i) {
        return ((i)*(i+1L))/2L;
    }

    boolean good(long l, long r, long t, long m) {
        if ( r > (long)1e6 ) return false;
        long sum_of_s = (r-l+1L)*a + b*(sum(r-1L) - sum(l-2L));
        long max_si = S(r);
        return sum_of_s <= m*t && t >= max_si;
    }

    public void solve(int testNum, InputReader in, PrintWriter out) {
        a = in.nextInt();
        b = in.nextInt();
        n = in.nextInt();

        for ( int i = 0; i < n; i++ ) {
            long l = in.nextInt();
            long t = in.nextInt();
            long m = in.nextInt();

            long ll = 0;
            long rr = (long)1e6;
            long answer = -1L;

            while ( ll <= rr ) {
                long mid = (ll+rr)/2L;

                if ( good(l, l+mid, t, m) ) {
                    answer = l+mid;
                    ll = mid + 1L;
                } else {
                    rr = mid - 1L;
                }
            }

            out.print(answer + "\n");
        }
    }
}

class InputReader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;

    public InputReader(InputStream stream) {
        reader = new BufferedReader(new InputStreamReader(stream), 32768);
        tokenizer = null;
    }

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

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

}