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
#include <bits/stdc++.h>
using namespace std;
long long a[2000005], s[2000005] = {0};
int main() {
  long long n, i, t, l, m, ql, qr, mid, A, B;
  scanf("%I64d%I64d%I64d", &A, &B, &n);
  a[1] = s[1] = A;
  for (i = 2; i <= 2000000; i++) {
    a[i] = a[i - 1] + B;
    s[i] = s[i - 1] + a[i];
  }
  while (n--) {
    scanf("%I64d%I64d%I64d", &l, &t, &m);
    ql = l;
    qr = 2000000;
    if (a[ql] > t) {
      printf("-1\n");
      continue;
    }
    while (ql + 1 < qr) {
      mid = (qr + ql) / 2;
      long long tmp = max(a[mid], (s[mid] - s[l - 1]) % m == 0
                                      ? (s[mid] - s[l - 1]) / m
                                      : ((s[mid] - s[l - 1]) / m + 1));
      if (tmp <= t)
        ql = mid;
      else
        qr = mid;
    }
    printf("%I64d\n", ql);
  }
  return 0;
}
