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;
#pragma comment(linker, "/STACK:255000000")
bool firstout = 1;
template <class T>
T &minn(T &a, T b) {
  if (b < a) a = b;
  return a;
}
template <class T>
T &maxx(T &a, T b) {
  if (a < b) a = b;
  return a;
}
int &madd(int &a, int b) {
  a += b;
  if (a >= 1000000007) a -= 1000000007;
  return a;
}
int &msub(int &a, int b) {
  a -= b;
  if (a < 0) a += 1000000007;
  return a;
}
int &mmult(int &a, int b) { return a = (long long)a * b % 1000000007; }
int mdiv(long long a, long long b, long long m) {
  a = (a % m + m) % m;
  b = (b % m + m) % m;
  if (a % b == 0) return a / b;
  return (a + m * mdiv(-a, m, b)) / b;
}
int n, m, q;
int A[1012];
int main() {
  int i, j, k;
  char c;
  int a, d;
  int b;
  int ts;
  for (ts = 1; scanf("%"
                     "d",
                     &(a)) > 0;
       ++ts) {
    scanf(
        "%"
        "d",
        &(b));
    scanf(
        "%"
        "d",
        &(q));
    for (; q--;) {
      int l, t, m;
      scanf(
          "%"
          "d",
          &(l));
      scanf(
          "%"
          "d",
          &(t));
      scanf(
          "%"
          "d",
          &(m));
      --l;
      i = l;
      if (a + (long long)l * b > t)
        printf(
            "%"
            "d",
            (-1)),
            printf("\n"), firstout = 1;
      else {
        j = (t - a) / b;
        if (j + 1 - l > m) {
          k = j;
          double x =
              (sqrt(((2. * a + 2. * b * l - b) * (2. * a + 2. * b * l - b)) +
                    8. * b * t * m) -
               (2. * a + 2. * b * l - b)) /
              2. / b;
          j = l + (int)x;
          for (; (long long)a * (j - i + 1) +
                     (long long)(j - i + 1) *
                         ((long long)l * b + (long long)(l + j - i) * b) / 2 >
                 (long long)t * m;)
            --j;
          minn(j, k);
        }
        printf((firstout) ? "%"
                            "d"
                          : " %"
                            "d",
               (j + 1)),
            firstout = 0;
        printf("\n"), firstout = 1;
      }
    }
  }
  return 0;
}
