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>
class segtree {
  void from_array(T* v, int idx, int left, int right) {
    if (left != right) {
      from_array(v, (2 * (idx) + 1), left, (((left) + (right)) / 2));
      from_array(v, (2 * (idx) + 2), (((left) + (right)) / 2) + 1, right);
      tree[idx] = max(tree[(2 * (idx) + 1)], tree[(2 * (idx) + 2)]);
    } else
      tree[idx] = v[left];
  }
  T read(int i, int j, int idx, int left, int right) {
    if (i <= left && right <= j) return tree[idx];
    if (j < left || right < i) return 0;
    return max(
        read(i, j, (2 * (idx) + 1), left, (((left) + (right)) / 2)),
        read(i, j, (2 * (idx) + 2), (((left) + (right)) / 2) + 1, right));
  }

 public:
  T* tree;
  int size;
  segtree() {}
  ~segtree() { delete[] tree; }
  segtree(int size) {
    this->size = size;
    tree = new T[3 * size];
  }
  inline void from_array(T array[]) { from_array(array, 0, 0, size - 1); }
  inline T read(int i, int j) { return read(i, j, 0, 0, size - 1); }
};
long long arr[6000005];
long long sum[6000005];
int findr(int l, int t, int m, segtree<long long>& maxtree) {
  int st = l, end = 6000005 - 1;
  int r = -1;
  long long suml = l ? sum[l - 1] : 0;
  while (st <= end) {
    int mid = (st + end) >> 1;
    long long maxv = maxtree.read(l, mid);
    long long sumv = sum[mid] - suml;
    if (maxv <= t && sumv <= (long long)t * m)
      st = mid + 1, r = mid + 1;
    else
      end = mid - 1;
  }
  return r;
}
int main() {
  int A, B, N;
  scanf(" %d %d %d", &A, &B, &N);
  segtree<long long> tree(6000005);
  for (int i = 0; i < 6000005; i++) {
    arr[i] = (long long)A + (long long)i * B;
    sum[i] = (i ? sum[i - 1] : 0LL) + arr[i];
  }
  tree.from_array(arr);
  for (int i = 0; i < N; i++) {
    int l, t, m;
    scanf(" %d %d %d", &l, &t, &m), l--;
    printf("%d\n", findr(l, t, m, tree));
  }
  return 0;
}
