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;


public class TavasAndKarafs {

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		
		String [] input = br.readLine().split(" ");
		
		long a = Long.parseLong(input[0]);
		long b = Long.parseLong(input[1]);
		long n = Long.parseLong(input[2]);
		
		while(n-- > 0) {
			input = br.readLine().split(" ");
			
			long l = Long.parseLong(input[0]);
			long t = Long.parseLong(input[1]);
			long m = Long.parseLong(input[2]);
			
			long s = a + (l - 1) * b;
			
			if(s > t || a > t) {
				System.out.println(-1);
				continue;
			}
			
			long low = l, high = (m * t - a) / b + 1; 
			
			while(high > low + 1) {
				long mid = (high + low) / 2;
				long aMid = (mid - 1) * b + a;
				long sum = (mid - l + 1) * (a + (l - 1) * b + aMid) / 2;
				if (sum <= m * t) {
					if(aMid <= t) {
						low = mid;
					}
					else {
						high = mid;
					}
				}
				else {
					high = mid;
				}
			}
			
			long aMid = (high - 1) * b + a;
			long sum = (high - l + 1) * (a + (l - 1) * b + aMid) / 2;
			
			long sol;
			if (sum <= m * t) {
				if(aMid <= t) {
					sol = high;
				}
				else {
					sol= low;
				}
			}
			else {
				sol = low;
			}

			System.out.println(sol);
		}
	}

}