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)

use proconio::input;

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

    let mut dp = vec![0u64; m];
    let mut i = (0..).peekable();
    while i.next().unwrap() < n {
        dp[x] += 1;
        x = x * x % m;
        if dp[x] == 2 {
            break;
        }
    }

    if i.peek().unwrap() < &(n - 1) {
        let per = dp.iter().filter(|&x| x == &2).count() as u64;
        let rem = n - i.peek().unwrap();
        let add = rem / per;
        for j in 0..m {
            if dp[j] == 2 {
                dp[j] += add;
            }
        }
        let ci = *i.peek().unwrap() + add * per;
        i = (ci..).peekable();

        while i.next().unwrap() < n {
            dp[x] += 1;
            x = x * x % m;
        }
    }

    let res = dp
        .into_iter()
        .enumerate()
        .map(|(i, x)| i as u64 * x)
        .sum::<u64>();
    println!("{}", res);
}