#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <cmath>
#include <queue>
#include <set>
#include <map>
#include <iomanip>
#include <fstream>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int mod = 1000000007;

int main() {
    ios::sync_with_stdio(false);
    int k, q;
    ll a[5003] = {};
    cin >> k >> q;
    for (int i = 0; i < k; i++) cin >> a[i];
    while (q--) {
        ll n, x, m, b[5003] = {}, c[5003] = {};
        cin >> n >> x >> m; n--;
        x %= m;
        for (int i = 1; i <= k; i++) b[i] = b[i - 1] + a[i - 1] % m;
        for (int i = 1; i <= k; i++) c[i] = c[i - 1] + (a[i - 1] % m == 0);
        ll s = b[k] * (n / k) + b[n % k] + x, t = c[k] * (n / k) + c[n % k];
        cout << n - s / m - t << '\n';
    }
}

### Prompt

Develop a solution in CPP to the problem described below:

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

### Response

```cpp
#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <cmath>
#include <queue>
#include <set>
#include <map>
#include <iomanip>
#include <fstream>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int mod = 1000000007;

int main() {
    ios::sync_with_stdio(false);
    int k, q;
    ll a[5003] = {};
    cin >> k >> q;
    for (int i = 0; i < k; i++) cin >> a[i];
    while (q--) {
        ll n, x, m, b[5003] = {}, c[5003] = {};
        cin >> n >> x >> m; n--;
        x %= m;
        for (int i = 1; i <= k; i++) b[i] = b[i - 1] + a[i - 1] % m;
        for (int i = 1; i <= k; i++) c[i] = c[i - 1] + (a[i - 1] % m == 0);
        ll s = b[k] * (n / k) + b[n % k] + x, t = c[k] * (n / k) + c[n % k];
        cout << n - s / m - t << '\n';
    }
}

```