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 PyPy3 (2.4.0) (judge verdict: Accepted)

import sys
input = sys.stdin.readline
k, q = map(int,input().split())
D = list(map(int,input().split()))
A = [list(map(int,input().split())) for i in range(q)]

for i in range(q):
    n = A[i][0]
    x = A[i][1]
    m = A[i][2]
    ans = 0
    B = [x%m]
    BP = [0] * (k+1)
    n -= 1
    for j in range(k):
        B.append((B[-1] + D[j]) % m)
    ans0 = 0
    for j in range(len(B)-1):
        if B[j+1] > B[j]:
            BP[j+1] = 1
            ans0 += 1
    # print(B,BP)
    if B[-1] != B[0]:
        if B[-1] > B[0]:
            ans1 = [ans0-1, ans0]
        elif B[-1] < B[0]:
            # print("sage")
            ans1 = [ans0, ans0+1]
        d = (B[-1] - B[0]) % m
        kn = n // k
        c = (B[0] + d * kn) % m
        sag = (B[0] + d * kn) // m
        # print(((B[0] + d * kn) - c) % m,"zero")
        ans = ans1[0] * sag + ans1[1] * (kn-sag)
        # print("kn",kn,"sag",sag,"c",c)
        for ii in range(n%k):
            cm = c
            # print("plus")
            c += D[ii]
            if c >= m:
                c %= m
            
            if c>cm:
                ans += 1    
        print(ans)
        
    else:
        # print("eq")
        print(ans0 * (n//k) + sum(BP[:n%k+1]))