//方針変えたら(たぶん)自明で草
#include <iostream>
#include <string>
#include <algorithm>
#include <functional>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <cmath>
#include <tuple>
#define rep(i, n) for(i = 0; i < n; i++)
#define int long long
using namespace std;

int k, q;
int d[5000];

int solve(int n, int x, int m) {
	int i;
	
	int s = x % m, ng = 0;
	rep(i, k) {
		int e = d[i] % m;
		int cnt;
		if (e == 0) { e = m; }
		if (n - 2 - i < 0) { cnt = 0; }
		else { cnt = (n - 2 - i) / k + 1; }
		s += e * cnt;
	}
	ng += s / m;
	return n - 1 - ng;
}

signed main() {
	int i;
	int n, x, m;

	cin >> k >> q;
	rep(i, k) cin >> d[i];
	
	rep(i, q) {
		cin >> n >> x >> m;
		int ans = solve(n, x, m);
		cout << ans << endl;
	}
	return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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
//方針変えたら(たぶん)自明で草
#include <iostream>
#include <string>
#include <algorithm>
#include <functional>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <cmath>
#include <tuple>
#define rep(i, n) for(i = 0; i < n; i++)
#define int long long
using namespace std;

int k, q;
int d[5000];

int solve(int n, int x, int m) {
	int i;
	
	int s = x % m, ng = 0;
	rep(i, k) {
		int e = d[i] % m;
		int cnt;
		if (e == 0) { e = m; }
		if (n - 2 - i < 0) { cnt = 0; }
		else { cnt = (n - 2 - i) / k + 1; }
		s += e * cnt;
	}
	ng += s / m;
	return n - 1 - ng;
}

signed main() {
	int i;
	int n, x, m;

	cin >> k >> q;
	rep(i, k) cin >> d[i];
	
	rep(i, q) {
		cin >> n >> x >> m;
		int ans = solve(n, x, m);
		cout << ans << endl;
	}
	return 0;
}
```