We have a sequence of k numbers: d_0,d_1,...,d_{k - 1}.

Process the following q queries in order:

* The i-th query contains three integers n_i, x_i, and m_i. Let a_0,a_1,...,a_{n_i - 1} be the following sequence of n_i numbers: \begin{eqnarray} a_j = \begin{cases} x_i & ( j = 0 ) \\\ a_{j - 1} + d_{(j - 1)~\textrm{mod}~k} & ( 0 < j \leq n_i - 1 ) \end{cases}\end{eqnarray} Print the number of j~(0 \leq j < n_i - 1) such that (a_j~\textrm{mod}~m_i) < (a_{j + 1}~\textrm{mod}~m_i).



Here (y~\textrm{mod}~z) denotes the remainder of y divided by z, for two integers y and z~(z > 0).

Constraints

* All values in input are integers.
* 1 \leq k, q \leq 5000
* 0 \leq d_i \leq 10^9
* 2 \leq n_i \leq 10^9
* 0 \leq x_i \leq 10^9
* 2 \leq m_i \leq 10^9

Input

Input is given from Standard Input in the following format:


k q
d_0 d_1 ... d_{k - 1}
n_1 x_1 m_1
n_2 x_2 m_2
:
n_q x_q m_q


Output

Print q lines.

The i-th line should contain the response to the i-th query.

Examples

Input

3 1
3 1 4
5 3 2


Output

1


Input

7 3
27 18 28 18 28 46 1000000000
1000000000 1 7
1000000000 2 10
1000000000 3 12


Output

224489796
214285714
559523809
#include <bits/stdc++.h>
using namespace std;

int main() {
  int k, q;
  cin >> k >> q;
  vector<int64_t> d(k), n(q), x(q), m(q);
  for (int i = 0; i < k; i++) cin >> d.at(i);
  for (int i = 0; i < q; i++) cin >> n.at(i) >> x.at(i) >> m.at(i);
  for (int i = 0; i < q; i++) {
    vector<int64_t> dm(k);
    int dm0 = 0;
    for (int j = 0; j < k; j++) {
      dm.at(j) = d.at(j) % m.at(i);
      if (dm.at(j) == 0) dm0++;
    }
    x.at(i) %= m.at(i);
    int64_t s = accumulate(dm.begin(), dm.end(), 0ll),
      sq = s / m.at(i), sr = s % m.at(i), l = (n.at(i) - 1) / k;
    int64_t ans = n.at(i) - 1, r = x.at(i);
    ans -= l * (sq + dm0);
    for (int j = 0; j < (n.at(i) - 1) % k; j++) {
      r += dm.at(j);
      if (dm.at(j) == 0) ans--;
    }
    ans -= (l * sr + r) / m.at(i);
    cout << ans << "\n";
  }
}
