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: Wrong Answer)

//#pragma GCC optimize("O3")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define trav(x,a) for (const auto& x: a)
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), x.end()
#define mem(a,v) memset((a), (v), sizeof (a))
#define enl printf("\n")
#define case(t) printf("Case #%d: ", (t))
#define ni(n) scanf("%d", &(n))
#define nl(n) scanf("%lld", &(n))
#define nai(a, n) for (int _i = 0; _i < (n); _i++) ni(a[_i])
#define nal(a, n) for (int _i = 0; _i < (n); _i++) nl(a[_i])
#define pri(n) printf("%d\n", (n))
#define prl(n) printf("%lld\n", (n))
#define pii pair<int, int>
#define pll pair<long long, long long>
#define vii vector<pii>
#define vll vector<pll>
#define vi vector<int>
#define vl vector<long long>
#define pb push_back
#define mp make_pair
#define st first
#define nd second
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef cc_hash_table<int,int,hash<int>> ht;
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> oset;
const double pi = acos(-1);
const int mod = 1e9 + 7;
const int inf = 1e9 + 7;
const int N = 1e6 + 5;
const double eps = 1e-9;
int d[N];

int main() {
  int k, q; scanf("%d %d", &k, &q);
  for (int i = 0; i < k; i++)
    scanf("%d", &d[i]);
  for (int i = 1; i <= q; i++) {
    int n, x, m; scanf("%d %d %d", &n, &x, &m);
    ll sm = 0;
    for (int j = 0; j < k; j++)
      sm += d[j] % m;
    x %= m;
    ll ans = 0;
    if (n <= k) {
      int a = x, z = x;
      for (int j = 1; j < n; j++) {
        z = z + (d[(j-1)%k] % m);
        int b = (a + d[(j - 1) % k]) % m;
        if (a < b)
          ans++;
        a = b;
      }
    } else {
      ll z = x + (n/k) * 1ll * sm;
      for (int j = 1; j < (n % k); j++)
        z += d[(j - 1) % k] % m;
      ans = z / m;
      for (int j = 0; j < k; j++) if (d[j] % m == 0)
        ans += (j+1 < (n % k)) + (n / k);
      ans = n - ans - 1;
    }
    printf("%lld\n", ans);
  }
  return 0;
}