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.*;
import java.io.*;

public class Round_535A {

    static BufferedReader br;
    static PrintWriter pr;
    static StringTokenizer st;

    public static void main (String[] args) throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        pr = new PrintWriter(new OutputStreamWriter(System.out));
        //br = new BufferedReader(new FileReader("in.txt"));
        //pr = new PrintWriter(new FileWriter("out.txt"));

        long a = readInt();
        long b = readInt();
        int n = readInt();
        for (int i = 0; i < n; i++) {
            int l = readInt();
            long t = readInt();
            long m = readInt();
            int lo = l-1;
            int hi = 1000010;
            while (lo <= hi) {
                int mid = lo + (hi - lo)/2;
                long max = a + (mid-2) * b;
                long largest = b*(mid-l-1);
                
                long sum = (a+(l-1)*b)*(mid-l) + largest*((mid-l)/2) + ((mid-l)%2 == 1 ? largest/2 : 0);
//              System.out.println((l) + " " + (mid-1) + " " + max + " " + sum + " " + largest);
                if (max <= t && sum <= m*t) {
                    lo = mid + 1;
                } else {
                    hi = mid - 1;
                }
            }
//          System.out.println(hi);
            if (hi <= l)
                System.out.println(-1);
            else
                System.out.println(hi-1);
        }
        
        pr.close();
    }

    static String next () throws IOException {
        while (st == null || !st.hasMoreTokens())
            st = new StringTokenizer(br.readLine().trim());
        return st.nextToken();
    }

    static long readLong () throws IOException {
        return Long.parseLong(next());
    }

    static int readInt () throws IOException {
        return Integer.parseInt(next());
    }

    static double readDouble () throws IOException {
        return Double.parseDouble(next());
    }

    static char readCharacter () throws IOException {
        return next().charAt(0);
    }

    static String readLine () throws IOException {
        return br.readLine().trim();
    }
}
