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 f[2000005];
int A, B, Q, i, L, l, r, mid;
int main() {
  scanf("%d%d%d", &A, &B, &Q);
  f[1] = A;
  for (i = 2; i <= 2000000; i++) f[i] = f[i - 1] + B;
  for (i = 1; i <= 2000000; i++) f[i] += f[i - 1];
  while (Q--) {
    scanf("%d%d%d", &L, &A, &B);
    l = L;
    r = 2000000;
    mid = (l + r) / 2;
    while (l <= r) {
      if (f[mid] - f[mid - 1] > A || f[mid] - f[L - 1] > 1ll * A * B) {
        r = mid - 1;
        mid = (l + r) / 2;
      } else {
        l = mid + 1;
        mid = (l + r) / 2;
      }
    }
    if (r < L)
      puts("-1");
    else
      printf("%d\n", r);
  }
  return 0;
}
