#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define mp(a, b) make_pair(a, b)
typedef pair<int, int> pii;
inline int lowbit(int x) { return (x & (-x));}
const int maxn = 2e5 + 10;
const ll mod = 1e9 + 7;
int d[maxn], a[maxn];
ll sum[maxn], tmp[maxn];
int main()
{
	int k, q;
	cin >> k >> q;
	for(int i = 0; i < k; ++i) cin >> d[i];
	for(int i = 1; i <= q; ++i)
	{
		int n, x, m;
		cin >> n >> x >> m;
		ll ret = x;
		for(int j = 0; j < k; ++j)
		{
			a[j] = d[j] % m;
			sum[j + 1] = sum[j] + a[j];
			tmp[j + 1] = tmp[j] + (a[j] == 0);
		}
		ret += sum[k] * ((n - 1) / k) + sum[(n - 1) % k];
		cout << n - 1 - (ret / m - x / m) - (tmp[k] * ((n - 1) / k) + tmp[(n - 1) % k]) << endl;
		 
	}
	return 0;
}
	
### Prompt

Generate a cpp solution to the following problem:

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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define mp(a, b) make_pair(a, b)
typedef pair<int, int> pii;
inline int lowbit(int x) { return (x & (-x));}
const int maxn = 2e5 + 10;
const ll mod = 1e9 + 7;
int d[maxn], a[maxn];
ll sum[maxn], tmp[maxn];
int main()
{
	int k, q;
	cin >> k >> q;
	for(int i = 0; i < k; ++i) cin >> d[i];
	for(int i = 1; i <= q; ++i)
	{
		int n, x, m;
		cin >> n >> x >> m;
		ll ret = x;
		for(int j = 0; j < k; ++j)
		{
			a[j] = d[j] % m;
			sum[j + 1] = sum[j] + a[j];
			tmp[j + 1] = tmp[j] + (a[j] == 0);
		}
		ret += sum[k] * ((n - 1) / k) + sum[(n - 1) % k];
		cout << n - 1 - (ret / m - x / m) - (tmp[k] * ((n - 1) / k) + tmp[(n - 1) % k]) << endl;
		 
	}
	return 0;
}
	
```