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>
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define rep(i,N) for(long long i = 0; i < (long long)(N); i++)
#define repr(i,N) for(long long i = (long long)(N) - 1; i >= 0; i--)
#define rep1(i,N) for(long long i = 1; i <= (long long)(N) ; i++)
#define repr1(i,N) for(long long i = (N) ; (long long)(i) > 0 ; i--)
#define each(x,v) for(auto& x : v)
#define all(v) (v).begin(),(v).end()
#define sz(v) ((int)(v).size())
#define ini(...) int __VA_ARGS__; in(__VA_ARGS__)
#define inl(...) long long __VA_ARGS__; in(__VA_ARGS__)
#define ins(...) string __VA_ARGS__; in(__VA_ARGS__)
using namespace std; void solve();
using ll = long long; template<class T = ll> using V = vector<T>;
using vi = V<int>; using vl = V<>; using vvi = V< V<int> >;
constexpr int inf = 1001001001; constexpr ll infLL = (1LL << 61) - 1;
struct IoSetupNya {IoSetupNya() { cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(15); cerr << fixed << setprecision(7);} } iosetupnya;
template<typename T, typename U> inline bool amin(T &x, U y) { return (y < x) ? (x = y, true) : false; }
template<typename T, typename U> inline bool amax(T &x, U y) { return (x < y) ? (x = y, true) : false; }
template<typename T, typename U> ostream& operator <<(ostream& os, const pair<T, U> &p) { os << p.first << " " << p.second; return os; }
template<typename T, typename U> istream& operator >>(istream& is, pair<T, U> &p) { is >> p.first >> p.second; return is; }
template<typename T> ostream& operator <<(ostream& os, const vector<T> &v) { int s = (int)v.size(); rep(i,s) os << (i ? " " : "") << v[i]; return os; }
template<typename T> istream& operator >>(istream& is, vector<T> &v) { for(auto &x : v) is >> x; return is; }
void in(){} template <typename T,class... U> void in(T &t,U &...u){ cin >> t; in(u...);}
void out(){cout << "\n";} template <typename T,class... U> void out(const T &t,const U &...u){ cout << t; if(sizeof...(u)) cout << " "; out(u...);}
template<typename T>void die(T x){out(x); exit(0);}
#ifdef NyaanDebug
  #include "NyaanDebug.h"
  #define trc(...) do { cerr << #__VA_ARGS__ << " = "; dbg_out(__VA_ARGS__);} while(0)
  #define trca(v,N) do { cerr << #v << " = "; array_out(v , N);} while(0)
  #define trcc(v) do { cerr << "name : " << #v << "\n"; int cnt = 0; each(x , v){cerr << (cnt++) << " : "; trc(x); } } while(0)
#else
  #define trc(...)
  #define trca(...)
  #define trcc(...)
  int main(){solve();}
#endif

#define inc(...) char __VA_ARGS__; in(__VA_ARGS__)
#define in2(s,t)   rep(i,sz(s)){in(s[i] , t[i]);}
#define in3(s,t,u) rep(i,sz(s)){in(s[i] , t[i] , u[i]);}
using vd = V<double>; using vs = V<string>; using vvl = V< V<> >;
template<typename T,typename U>ll ceil(T a,U b){return (a + b - 1) / b;}

using P = pair<ll,ll>; using vp = V<P>;
constexpr int MOD = /**/ 1000000007; //*/ 998244353;
//////////////////

void solve(){
  inl(K,Q);
  vl d(K); in(d);

  rep(_,Q){
    inl(N,X,M);
    vl a = d;
    each(x,a) {if( (x %= M) == 0) x = M;}
    ll s = 0; each(x,a) s += x;
    trc(a);
    // 全部で (N - 1) / K 周と　(N - 1) % K 回
    
    ll nya = s * ((N - 1) / K);
    rep(i , (N - 1) % K) nya += a[i];
    trc(nya);
    // X から　X + nyaまで何回0をまたいだ？
    ll ans = (nya + X) / M - X / M;
    ans = N - 1 - ans;
    out(ans);
    
  }

}