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::*;

fn run() {
    input! {
        mut n: usize,
        x: usize,
        m: usize,
    }
    let mut dp = vec![(0, 0); m];
    for (i, dp) in dp.iter_mut().enumerate() {
        *dp = (i * i % m, i as u64);
    }
    type T = (usize, u64);
    let mul = |a: &[T], b: &[T]| -> Vec<T> {
        let mut c = vec![(0, 0); m];
        for (c, a) in c.iter_mut().zip(a.iter()) {
            let b = b[a.0];
            *c = (b.0, b.1 + a.1);
        }
        c
    };
    let mut t = (0..m).map(|p| (p, 0)).collect::<Vec<_>>();
    let mut s = dp;
    while n > 0 {
        if n & 1 == 1 {
            t = mul(&t, &s);
        }
        s = mul(&s, &s);
        n >>= 1;
    }
    let ans = t[x].1;
    println!("{}", ans);
}

fn main() {
    run();
}