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: Wrong Answer)

use proconio::input;
use std::cmp;

fn main() {
    input!{
        n: usize,
        x: usize,
        m: usize,
    }
    // let n = 1000;
    // let x = 2;
    // let m = 449;
    let mut ans = x;
    let mut a = vec![0; m];
    a[0] = x;
    for i in 1..cmp::min(n, m) {
        a[i] = a[i-1] * a[i-1] % m;
        ans += a[i];
//        println!("{}", a[i]);
        let mut flag = 0;
        for j in 0..i {
            if a[j] == a[i] {
                flag = j;
            }
        }
        if flag != 0 {
        // println!("{}", (n-i) / (i-flag));
            for j in flag..i {
                ans += a[j] * ((n-i) / (i-flag));
            }
            let mut x = a[i];
            let mut y;
            for _ in i+((n-i)/(i-flag))*(i-flag)+1..n {
                y = x * x % m;
                ans += y;
        // println!("{}", y);
                x = y;
            }
            break;
        }
    }
    println!("{}", ans);
}