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;
use proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;

// #[cfg_attr(cargo_equip, cargo_equip::equip)]
// use ::__kyopro_lib::{};

fn main() {
    input! {
        n: u64,
        x: usize,
        m: usize,
    }

    let mx_k = 50;
    // (next, sum)
    let mut t = vec![vec![(0, 0); mx_k]; m];

    for i in 0..m {
        t[i][0] = (i * i % m, i);
    }
    for k in 0..mx_k - 1 {
        for i in 0..m {
            let x = t[i][k];
            let y = t[x.0][k];
            t[i][k + 1] = (y.0, y.1 + x.1);
        }
    }

    // println!("{:?}", t);

    let mut res = 0;
    let mut x = x;
    for k in 0..mx_k {
        if n >> k & 1 == 1 {
            res += t[x][k].1 as u64;
            x = t[x][k].0;
        }
    }

    println!("{}", res);
}