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)

macro_rules! read_line_to_tuple {
    ( $( $t:ty ),* ) => {{
        let mut input = String::new();
        std::io::stdin().read_line(&mut input).unwrap();
        let mut iter = input.split_whitespace();
        ( $( iter.next().unwrap().parse::<$t>().unwrap() ),* )
    }};
}

use std::collections::HashMap;
use std::cmp::min;

fn main() {
    let (n, x, m) = read_line_to_tuple!(usize, usize, usize);

    let mut map: HashMap<usize, usize> = HashMap::new();
    let mut v = Vec::with_capacity(m);
    let mut a = x;
    for i in 0..(min(n, m)) {
        if map.contains_key(&a) {
            break;
        } else {
            map.insert(a, i);
            v.push(a);
            a = a * a % m;
        }
    }

    let mut ans = 0;
    if n <= v.len() {
        for i in 0..n {
            ans += v[i];
        }
        return println!("{}", ans);
    }
    let left = map[&a];
    let right = v.len();
    for i in 0..left {
        ans += v[i];
    }
    for k in left..(right - 1) {
        v[k + 1] += v[k];
    }
    let q = (n - left) / (right - left);
    let r = (n - left) % (right - left);
    ans += q * v[right - 1] + if r == 0 { 0 } else { v[left + r - 1] };
    println!("{}", ans);
}