Problem:
Score : 500 points

Problem Statement

Let us denote by f(x, m) the remainder of the Euclidean division of x by m.

Let A be the sequence that is defined by the initial value A_1=X and the recurrence relation A_{n+1} = f(A_n^2, M).
Find \displaystyle{\sum_{i=1}^N A_i}.

Constraints

1 \leq N \leq 10^{10}

0 \leq X < M \leq 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N X M

Output

Print \displaystyle{\sum_{i=1}^N A_i}.

Sample Input 1

6 2 1001

Sample Output 1

1369

The sequence A begins 2,4,16,256,471,620,\ldots Therefore, the answer is 2+4+16+256+471+620=1369.

Sample Input 2

1000 2 16

Sample Output 2

6

The sequence A begins 2,4,0,0,\ldots Therefore, the answer is 6.

Sample Input 3

10000000000 10 99959

Sample Output 3

492443256176507

Solution in Rust (1.42.0) (judge verdict: Accepted)

#[allow(unused_imports)]
use {
    itertools::Itertools,
    proconio::{fastout, input, marker::*},
    std::cmp::*,
    std::collections::*,
    std::io::Write,
    std::ops::*,
};

#[allow(unused_macros)]
macro_rules! dbg { ($($e:expr),*) => { #[cfg(debug_assertions)] $({ let (e, mut err) = (stringify!($e), std::io::stderr()); writeln!(err, "{} = {:?}", e, $e).unwrap() })* }; }

#[fastout]
fn main() {
    input! {
        n: usize,
        x: usize,
        m: usize,
    }
    let mut r = x;
    let mut l = vec![x];
    let mut set = BTreeSet::new();
    set.insert(x);
    loop {
        r = (r * r) % m;
        l.push(r);
        if set.contains(&r) {
            break;
        }
        set.insert(r);
    }
    let l_len = l.len();
    let mut ans = 0;
    let mut k1 = 0;
    for i in 0..l_len {
        if l[i] == l[l_len - 1] {
            k1 = i;
            break;
        }
        ans += l[i];
    }
    l.pop();
    let k2 = l_len - k1 - 1;
    let mut ll = l[k1..]
        .iter()
        .scan(0, |arr, a| {
            *arr += a;
            Some(*arr)
        })
        .collect::<Vec<_>>();
    ll.insert(0, 0);
    if n > k1 {
        ans += ll[k2] * ((n - k1) / k2) + ll[(n - k1) % k2];
    }
    println!("{}", ans);
}