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)

#![allow(dead_code)]
#[allow(unused_imports)]
use proconio::{
    fastout, input,
    marker::{Bytes, Chars, Isize1, Usize1},
};
#[fastout]
fn main() {
    input!(n: usize, x: usize, m: usize);

    let mut ans = vec![x; 1];

    for i in 0..n + 1 {
        let buff = ans[i].pow(2) % m;
        if ans.contains(&buff) {
            let pos = ans.iter().position(|x| *x == buff).unwrap();
            let ans_len = ans.len();
            let wa = ans_len - pos;

            let count = (n - ans_len) / wa;
            let count_m = (n - ans_len) % wa;

            let mut sum = ans.iter().sum::<usize>();

            sum += ans.iter().skip(ans_len - wa).sum::<usize>() * count;
            sum += ans.iter().skip(ans_len - wa).take(count_m).sum::<usize>();
            println!("{}", sum);
            return;
        }

        ans.push(buff);
    }

    println!("{}", ans.iter().sum::<usize>());
}