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 Rust (1.42.0) (judge verdict: Accepted)

use std::io::{BufRead, Write};

fn main () {
    let stdin = std::io::stdin();
    let stdout = std::io::stdout();
    let mut reader = std::io::BufReader::new(stdin.lock());
    let mut writer = std::io::BufWriter::new(stdout.lock());

    let (k, q): (usize, usize) = {
        let mut buf = String::new();
        reader.read_line(&mut buf).unwrap();
        let mut iter = buf.split_whitespace();
        (
            iter.next().unwrap().parse().unwrap(),
            iter.next().unwrap().parse().unwrap(),
        )
    };

    let d: Vec<usize> = {
        let mut buf = String::new();
        reader.read_line(&mut buf).unwrap();
        let iter = buf.split_whitespace();
        iter.map(|x| x.parse().unwrap()).collect()
    };

    for _ in 0..q {
        let (n, x, m): (usize, usize, usize) = {
            let mut buf = String::new();
            reader.read_line(&mut buf).unwrap();
            let mut iter = buf.split_whitespace();
            (
                iter.next().unwrap().parse().unwrap(),
                iter.next().unwrap().parse().unwrap(),
                iter.next().unwrap().parse().unwrap(),
            )
        };
        let x = x % m;
    
        let mut res = 0;
        let (mut quo, mut rem) = (0, x);
        for i in 0..(std::cmp::min(n - 1, k)) {
            let c = 1 + ((n - 1) - (i + 1)) / k;
            let d = d[i] % m;
            if d == 0 {
                res += c;
            } else {
                quo += (rem + d * c) / m;
                rem = (rem + d * c) % m;
            }
        }

        let ans = (n - 1) - (res + quo - x / m);
        writeln!(writer, "{}", ans).unwrap();
    }
}