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.*;
public class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        
        
          int n;
                A = sc.nextInt();B=+sc.nextInt();n=sc.nextInt();
                while(n--!=0){
                l = sc.nextInt();t=sc.nextInt();m=sc.nextInt();
                     System.out.println(bins());
                }
        }
    static long l,t,m;
    static long A,B;
    static long MAXN = (long)1E9;
       static long bins(){
           long s = 1,e = (int)1e9;
           long ans=-1;
           while(s<=e){
                          long mid = (e+s)/2;

//               System.out.print(s+" "+e+" "+mid+" "+sum(mid));
               
               long iom = A+(mid-1)*B;
//               System.out.println(" "+iom);
               if(sum(mid)<=m*t && iom <= t ) {ans = Math.max(ans,mid);s=mid+1;}
               if(iom > t){e=mid-1;}
               else if(sum(mid)  < m*t ){
                   s=mid+1;
               }
               else e=mid-1;
           }
           if(ans<l) return -1;
           return ans;
       }
       static long sum(long r){
//           System.out.println("sum");
//           System.out.println((r-l+1)*A+r*(r-1)*B/2-(l-1)*(l-2)*B/2);
        return (r-l+1)*A+r*(r-1)*B/2-(l-1)*(l-2)*B/2;
       }
}