Problem:
Score : 600 points

Problem Statement

Given is a positive integer N.

We will choose an integer K between 2 and N (inclusive), then we will repeat the operation below until N becomes less than K.

Operation: if K divides N, replace N with N/K; otherwise, replace N with N-K.

In how many choices of K will N become 1 in the end?

Constraints

2 \leq N \leq 10^{12}

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of choices of K in which N becomes 1 in the end.

Sample Input 1

6

Sample Output 1

3

There are three choices of K in which N becomes 1 in the end: 2, 5, and 6.

In each of these choices, N will change as follows:

When K=2: 6 \to 3 \to 1

When K=5: 6 \to 1

When K=6: 6 \to 1

Sample Input 2

3141

Sample Output 2

13

Sample Input 3

314159265358

Sample Output 3

9

Solution in Rust (1.42.0) (judge verdict: Wrong Answer)

use proconio::{fastout, input};

#[fastout]
fn main() {
    input! {in_n: i64};
    let mut count = 0;

    for p in 2..=in_n {
        if p * p > in_n {
            break;
        }

        if in_n % p != 0 {
            continue;
        }

        let q = in_n / p;
        let mut n = in_n;
        let mut m = in_n;

        while n % p == 0 {
            n /= p;
        }

        while m % q == 0 {
            m /= q;
        }

        if n % p == 1 {
            count += 1;
        }

        if m != q && p % q == 1 {
            count += q;
        }
    }

    for p in 2..=in_n {
        if p * p > in_n {
            break;
        }

        if in_n % p != 1 {
            continue;
        }

        let q = (in_n - 1) / p;

        if p == q {
            count += 1;
        } else {
            count += 2;
        }
    }

    println!("{}", count + 2);
}