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)

#![allow(unused_imports, non_snake_case)]
use proconio::{
    input, fastout,
    marker::{Chars, Bytes, Usize1}
};

#[fastout]
fn main() {
    input!{
        N:usize,
        X:usize,
        M:usize,
    }
    let mut g = vec![0; M];
    let mut cur = X;
    let mut cursum = X;

    for i in 1..M*10 {
        if i == N {
            println!("{}", cursum);
            break;
        }
        if g[cur] == 0 {
            g[cur] = i;
        } else {
            let rot = i - g[cur];
            let mut rotsum = cursum;
            for j in 0..M {
                if 0 < g[j] && g[j] <= g[cur] {
                    rotsum -= j;
                }
            }
            let rest = N - i;
            cursum += rotsum * (rest / rot);
            for _ in 0..rest%rot {
                cur = cur * cur % M;
                cursum += cur;
            }
            println!("{}", cursum);
            break;
        }
        cur = cur * cur % M;
        cursum += cur;
    }
}