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.

<image>

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.*;
public class _535C_Tavas_and_Karafs {
	static long mor,S[];
	static int I[],l,t,m,A,B,li,ls,r,h;
	public static void Calcula(){
		if(li<=ls){
			int i = (li+ls)/2;
			if(S[i]==0){S[i]=B;S[i]*=i;S[i]*=(i-1);S[i]/=2;S[i]+=(long)(i*A);}
			long s = S[i]-S[l]+h;
			if(s!=mor){
				if(s<mor){
					long e=(long)(A+i*B); 
					s+=e;
					if(s<=mor && e<=t){
						li=i+1;Calcula();
					}
					else r=i;
				}else{
					ls=i;Calcula();
				}
			}else r=i;
		}else r=-1;
	}
	public static void main(String[] args){
       Scanner leer = new Scanner(System.in);
       A = leer.nextInt();
       B = leer.nextInt();
       int n = leer.nextInt();
       S= new long[1000001];
       I= new int[1000001];
       for (int i = 0; i < n; i++){
		   l = leer.nextInt();
		   t = leer.nextInt();
		   m = leer.nextInt();
		   mor=t;mor*=m; 
		   li=l;
		   h=A+(l-1)*B;
		   if(I[t]==0){
	    	   I[t]=(t-A)/B+1;
	       }
		   ls=I[t];
		   if(S[l]==0){S[l]=B;S[l]*=l;S[l]*=(l-1);S[l]/=2;S[l]+=(long)(l*A);}
		   Calcula();
	       System.out.println(r);
       }
	}
}