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(non_snake_case)]
use std::collections::HashMap;
use std::io::*;
use std::str::FromStr;

fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

fn main() {
    let n: usize = read();
    let mut x: usize = read();
    let m: usize = read();
    let mut counter = 0;
    let mut ans = 0;
    let mut surplus: HashMap<usize, usize> = HashMap::new();
    let mut sums: Vec<usize> = Vec::new();
    sums.push(0);
    // 方針 あまりの繰り返しを探して計算
    while counter != n {
        let val = surplus.get(&x);
        ans += x;
        sums.push(ans);
        counter += 1;
        match val {
            None => {
                surplus.insert(x, counter);
            }
            Some(&i) => {
                let loopVal = (n - counter) / (counter - i);
                let div = (n - counter) % (counter - i);
                ans += loopVal * (ans - sums[i]);
                ans += sums[div + i] - sums[i];
                break;
            }
        }
        x = (x * x) % m;
        if x == 0 {
            break;
        }
    }
    print!("{}", ans);
}