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 {
    static long A, B;

    public static void main(String[] args) throws IOException {
        BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
        String[] line = bf.readLine().split(" ");
        A = Integer.parseInt(line[0]);
        B = Integer.parseInt(line[1]);
        int n = Integer.parseInt(line[2]);
        int l, t, m;
        long lo, hi, mid;
        for (int i = 0; i < n; i++) {
            line = bf.readLine().split(" ");
            l = Integer.parseInt(line[0]);
            t = Integer.parseInt(line[1]);
            m = Integer.parseInt(line[2]);
            lo = l;
            hi = 10000000;
            while (lo + 1 < hi) {
                mid = (lo + hi) >> 1;
                if (check(mid, t, m, l)) {
                    lo = mid;
                } else {
                    hi = mid;
                }
            }
            if (check(lo, t, m, l))
                System.out.println(lo);
            else
                System.out.println(-1);
        }
    }

    static boolean check(long r, int t, int m, int l) {
        long nth = A + B * (r - 1);
        if (t < nth)
            return false;
        long fn = f(r) - f(l - 1);
        long have = ((long) t) * ((long) m);
        return have >= fn;
    }

    static long f(long n) {
        if (n <= 0)
            return 0;
        return A * n + B * ((n * (n - 1)) / 2);
    }
}