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.*;
import java.util.*;

public class Main {
    public static void main(String[] args) {
        long A, B; int n;
        Scanner in = new Scanner(System.in);
        A = in.nextLong(); B = in.nextLong(); n = in.nextInt();
        while ((n--) != 0) {
            long L;             long t, m;
            L = in.nextLong(); t = in.nextLong(); m = in.nextLong();
            if (A + (L - 1) * B > t) {
                System.out.println(-1);
                continue;
            }
            long tl = L, tr = (t - A) / B + 2;
            while (tr - tl > 1) {
                long mid = (tl + tr) / 2;
                if ((L - 1 + mid - 1) * (mid - L + 1) / 2 * B + (mid - L + 1) * A <= t * m) tl = mid;
                else tr = mid;
            }
            System.out.println(tl);
        }
    }
}