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;
use proconio::marker::Chars;
#[allow(unused_imports)]
use std::cmp::{max, min};
use std::collections::HashMap;
#[allow(unused)]
const ALPHA_SMALL: [char; 26] = [
    'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's',
    't', 'u', 'v', 'w', 'x', 'y', 'z',
];
#[allow(unused)]
const ALPHA: [char; 26] = [
    'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S',
    'T', 'U', 'V', 'W', 'X', 'Y', 'Z',
];
macro_rules! maxs {
    ($x:expr, $($y:expr), *) => {{
        let mut ret = $x;
        $(
            ret = max(ret, $y);
        )*
        ret
    }};
}

fn main() {
    input!(N: usize, X: usize, M: usize);
    let mut set = HashMap::new();
    let mut sum = vec![0; M + 1];
    let mut now = X;
    let mut L = (0, 0);
    sum[0] = X;
    if X == 0 {
        println!("0");
        return;
    }
    for i in 1..=M + 10 {
        let next = now * now % M;
        sum[i] = sum[i - 1] + next;
        if let Some(&h) = set.get(&(now, next)) {
            L = (h + 1, i);
            break;
        }
        set.insert((now, next), i);
        now = next;
    }
    let (h, i) = L;
    let l = i - h + 1;
    let sumL = sum[i] - sum[h - 1];
    let ans = sumL * ((N - h) / l) + sum[(N - h) % l + h - 1];
    println!("{}", ans);
}