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 Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		long A = sc.nextLong();
		long B = sc.nextLong();
		int n = sc.nextInt();
		for(int i=0;i<n;i++){
			long l = sc.nextLong();
			long t = sc.nextLong();
			long m = sc.nextLong();
			long sl = A + ((l - 1) * B);
			//System.out.println("s = " + s + " m = " + m + " t = " + t);
			if(t<sl){
				System.out.println(-1);
			}
			else{
				double a = B;
				double b = B + (2 * sl);
				double c = (2 * sl) - (2 * m * t);
				double x = ((-1 * b) + Math.sqrt((b * b) - (4 * a * c))) / (2 * a);
				long y = (long)x;
				long r = l + y;
				long sr = A + ((r - 1) * B);
				if(t>=sr){
					System.out.println(r);
				}
				else{
					long ans = ((t - A) / B) + 1;
					System.out.println(ans);
				}
			}
		}
	}
}
