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 math

A, B, n = map(int, input().split())
lCurKarafs = 0; lSum = 0; aa = 0; bb = 0; c = 0; D = 0; sol = 0; tth = 0;
while(n > 0):
    l, t, m = map(int, input().split())
    lCurKarafs = A + (l - 1) * B;
    #print(str(lCurKarafs) + " " + str(t) + " " + str(m))
    lSum = m * t;
    if(lCurKarafs > t):
        print(-1)
    elif(lCurKarafs == t):
        print(l)
    else:
        tth = (t - lCurKarafs + B) // B;
        aa = B; bb = 2 * lCurKarafs - B; c = (-2) * lSum;
        D = bb * bb - 4 * aa * c;
        #D = math.sqrt(D);
        sol = int((math.sqrt(D) - bb) / (2 * aa));
        sol = min(tth, sol);
        sol = l + sol - 1;
        print(sol)
    n -= 1