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: Wrong Answer)

use proconio::{fastout, input};
#[allow(unused_mut)]
#[fastout]
fn main() {
    input! {
        n: usize,
        x: usize,
        m: usize
    }
    let mut res = 0;
    let mut pattern = vec![x];
    let mut before = x;
    let mut firstpat = 0usize;
    let mut stopped = false;
    for _ in 0..n - 1 {
        let next = before * before;
        let next = next % m;
        if pattern.contains(&next) {
            firstpat = next;
            stopped = true;
            break;
        }
        res += next;
        pattern.push(next);
        before = next;
    }
    //println!("{:?}", pattern);
    if stopped {
        let firstindex = pattern.iter().position(|&i| i == firstpat).unwrap();
        //println!("{:?}", pattern);
        //ループ部分だけ取り出す
        let pattern: Vec<usize> = pattern.into_iter().skip(firstindex).collect();
        //println!("{:?}", pattern);

        let len = pattern.len();
        let sum: usize = pattern.iter().sum();
        res += (n / len) * sum;
        let s: usize = pattern.iter().take(n % len).sum();
        res += s;
    }
    println!("{}", res);
}