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;
template <class T>
inline bool rd(T &ret) {
  char c;
  int sgn;
  if (c = getchar(), c == EOF) return 0;
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  ret *= sgn;
  return 1;
}
template <class T>
inline void pt(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) pt(x / 10);
  putchar(x % 10 + '0');
}
const int N = 1e6 + 10;
const int mod = 1e9 + 7;
int n;
long long A, B;
long long l, t, m, mx;
long long siz(long long x) { return A + (x - 1) * B; }
long long work() {
  if (siz(l) - t > 0) return -1;
  long long L = l, R = mx, ans = l;
  while (L <= R) {
    long long mid = (L + R) >> 1;
    long long len = mid - l + 1;
    long long all = (siz(l) + siz(mid)) * len / 2;
    long long M = min(m, len);
    if (all > t * M || siz(mid) > t)
      R = mid - 1;
    else {
      ans = mid;
      L = mid + 1;
    }
  }
  return ans;
}
int main() {
  while (cin >> A >> B >> n) {
    mx = (1e6 - A) / B;
    mx++;
    while (n--) {
      rd(l);
      rd(t);
      rd(m);
      pt(work());
      puts("");
    }
  }
  return 0;
}
