Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3 1
3 1 4
5 3 2

Sample Output 1

1

For the first query, the sequence {a_j} will be 3,6,7,11,14.

(a_0~\textrm{mod}~2) > (a_1~\textrm{mod}~2)

(a_1~\textrm{mod}~2) < (a_2~\textrm{mod}~2)

(a_2~\textrm{mod}~2) = (a_3~\textrm{mod}~2)

(a_3~\textrm{mod}~2) > (a_4~\textrm{mod}~2)

Thus, the response to this query should be 1.

Sample Input 2

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

Sample Output 2

224489796
214285714
559523809

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
#define REP(i,n) for(int i=0;i<n;++i)
#define REP1(i,n) for(int i=1;i<=n;++i)
#define SZ(i) int(i.size())
#define eb emplace_back
#define ALL(i) i.begin(),i.end()
#define X first
#define Y second
#ifdef tmd
#define IOS()
#define debug(...) fprintf(stderr,"#%d: %s = ",__LINE__,#__VA_ARGS__),_do(__VA_ARGS__);
template<typename T> void _do(T &&x){cerr<<x<<endl;}
template<typename T, typename ...S> void _do(T &&x, S &&...y){cerr<<x<<", ";_do(y...);}
template<typename It> ostream& _printRng(ostream &os,It bg,It ed)
{
    os<<"{";
    for(It it=bg;it!=ed;it++) {
        os<<(it==bg?"":",")<<*it;
    }
    os<<"}";
    return os;
}
template<typename T> ostream &operator << (ostream &os,vector<T> &v){return _printRng(os,v.begin(), v.end());}
template<typename T> void pary(T bg, T ed){_printRng(cerr,bg,ed);cerr<<endl;}
#else
#define IOS() ios_base::sync_with_stdio(0);cin.tie(0);
#define endl '\n'
#define debug(...)
#define pary(...)
#endif

const int MAXN = 100005;
const ll MOD = 1000000007;

int k, q;
vector<int> d;

int solve (int n, int x, int m) {
    vector<int> dm;
    dm.reserve(k);

    for (auto v : d) {
        dm.eb(v % m);
    }

    x %= m;

    ll zs = 0;
    ll sum = x;
    for (int i=0; i<min(SZ(dm), n-1); i++) {
        int cnt = (n - i - 2) / k + 1;
        sum += 1LL * cnt * dm[i];
        debug(cnt, dm[i]);
        zs += (dm[i] == 0) * cnt;
    }
    debug(sum);

    return n - sum/m - zs - 1;
}
/*********************GoodLuck***********************/
int main () {
    IOS();

    cin >> k >> q;

    d.resize(k);
    REP (i, k) {
        cin >> d[i];
    }

    while (q--) {
        int n, x, m;
        cin >> n >> x >> m;
        cout << solve(n,x,m) << endl;
    }
}