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 std::collections::HashMap;
fn main() {
    input!(n: i128, x: i128, m: i128);
    let mut a = x;
    let mut res = x;
    let mut hashmap: HashMap<i128, i128> = HashMap::new();
    for i in 1..n {
        if hashmap.contains_key(&a) {
            let start = a;
            let mut next = a;
            let mut count = 1;
            let mut sum = a;
            loop {
                next = hashmap[&next];
                if next == start { break; }
                sum += next;
                count += 1;
            }
            let t = (n - i) / count;
            res += t * sum;
            next = hashmap[&a];
            for _ in 0..(n - (t * count + i)){
                res += next;
                next = hashmap[&next];
            }
            println!("{}", res);
            return;
        } else {
            let last_value = a;
            a = a * a % m;
            hashmap.insert(last_value, a);
            res += a;
        }
    }
    println!("{}", res);
}