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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
using namespace std;
int main() {
  int a, b, n, i, j;
  scanf("%d %d %d", &a, &b, &n);
  while (n--) {
    int l, t, m;
    scanf("%d %d %d", &l, &t, &m);
    int L = l - 1, R = 1000005;
    while (R - L > 1) {
      int mid = (R + L) / 2;
      bool ok = 1;
      long long num = a + (mid - 1) * 1ll * b;
      if (num > t) ok = 0;
      long long numl = a + (l - 1) * 1ll * b;
      long long sumr = (mid * (a + num)) / 2;
      long long suml = (l * (a + numl)) / 2;
      long long sum = sumr - suml + numl;
      if (m * 1ll * t < sum) ok = 0;
      if (ok)
        L = mid;
      else
        R = mid;
    }
    if (L == l - 1)
      printf("-1\n");
    else
      printf("%d\n", L);
  }
  return 0;
}
