#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 7;
const int mod = 1e9 + 7;
typedef long long ll;
int n, m, x;

ll ar[maxn], mid[maxn];
int main()
{
    int k, q;
    scanf("%d%d", &k, &q);
    for(int i=0;i<k;i++)scanf("%d", ar + i);
    while(q--){
        scanf("%d%d%d", &n, &x, &m);
        x %= m;
        ll s = 0;
        for(int i=0;i<k;i++){
            mid[i] = ar[i] % m;
            if(mid[i] == 0)mid[i] = m;
            s += mid[i];
        }
        ll a = s / m, b = s % m;
        a *= (n - 1) / k;
        a += b * ((n - 1) / k) / m;
        s = x + b * ((n - 1) / k) % m;
        //a += (n - 1) / k;
        for(int i=0;i<((n - 1) % k);i++)s += mid[i];
        a += s / m;
        printf("%lld\n", n - 1 - a);
    }
    return 0;
}

### Prompt

Please provide a Cpp coded solution 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 <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 7;
const int mod = 1e9 + 7;
typedef long long ll;
int n, m, x;

ll ar[maxn], mid[maxn];
int main()
{
    int k, q;
    scanf("%d%d", &k, &q);
    for(int i=0;i<k;i++)scanf("%d", ar + i);
    while(q--){
        scanf("%d%d%d", &n, &x, &m);
        x %= m;
        ll s = 0;
        for(int i=0;i<k;i++){
            mid[i] = ar[i] % m;
            if(mid[i] == 0)mid[i] = m;
            s += mid[i];
        }
        ll a = s / m, b = s % m;
        a *= (n - 1) / k;
        a += b * ((n - 1) / k) / m;
        s = x + b * ((n - 1) / k) % m;
        //a += (n - 1) / k;
        for(int i=0;i<((n - 1) % k);i++)s += mid[i];
        a += s / m;
        printf("%lld\n", n - 1 - a);
    }
    return 0;
}

```