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;

fn main() {
    input! {
        n: usize,
        x: usize,
        m: usize,
    }

    let ans = {
        let mut count = 0;
        let mut vec = vec![0; m+1];
        vec[x] = 0;
        let mut inv = vec![0; m+1];
        inv[0] = x;
        let mut next = x.pow(2) % m;
        while vec[next] == 0 {
            count += 1;
            inv[count] = next;
            vec[next] = count;
            next = next.pow(2) % m;
        }
        // vec[next]が最初のループ点の出現場所

        let mut ans = 0;
        for i in 0..n.min(vec[next]) {
            ans += inv[i]
        }
        if n > vec[next] {
            let amari = n-vec[next];
            let shuki = count-vec[next]+1;
            let mut aaa = 0;
            for i in 0..shuki {
                aaa += inv[vec[next]+i];
            }
            ans += (amari/shuki)*aaa;
            for i in 0..amari%shuki {
                ans += inv[vec[next]+i];
            }
        }
        ans
    };

    println!("{}",ans)
}