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 max_d = 50;
    let mut next = vec![vec![0; m]; max_d + 1];
    let mut sum = vec![vec![0u128; m]; max_d + 1];

    for r in 0..m {
        next[0][r] = r * r % m;
        sum[0][r] = r as u128;
    }

    for p in 0..max_d {
        for r in 0..m {
            //一つ上のマス(2^p先の値)が指してる列のp行目(2^p先の値)
            next[p + 1][r] = next[p][next[p][r]];
            sum[p + 1][r] = sum[p][r] + sum[p][next[p][r]];
        }
    }

    let mut ans = 0;
    let mut cur = x;
    for p in (0..=max_d).rev() {
        if (n & (1usize << p)) != 0 {
            ans += sum[p][cur];
            cur = next[p][cur];
        }
    }
    println!("{}", ans);
}