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 main() {
    input! {
        (n,x,m):(i64,i64,i64),
    };
    let mut n = n;
    let mut x = x;

    let mut id = vec![-1 as i64; m as usize];
    let mut len: i64 = 0;
    let mut total: i64 = 0;
    let mut a = vec![];

    while id[x as usize] == -1 {
        a.push(x);
        id[x as usize] = len;
        len += 1;
        total += x;
        x = x * x % m;
    }

    let c = len - id[x as usize];
    let mut s: i64 = 0;
    for i in 0..c {
        s += a[(len - i - 1) as usize];
    }

    let mut ans = 0;
    if n <= len {
        for i in 0..n {
            ans += a[i as usize];
        }
    } else {
        ans += total;
        n -= len;
        ans += s * (n / c);
        n %= c;
        for i in 0..n {
            ans += a[(id[x as usize] + i) as usize];
        }
    }
    println!("{}", ans);
}