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: Time Limit Exceeded)

use proconio::*;
use std::collections::*;

#[fastout]
fn main() {
    input! {
        n: usize,
    }

    let mut divs = vec![];
    let mut ddivs = vec![];
    for d in 2.. {
        if n % d == 0 {
            divs.push(d);
            divs.push(n / d);
        }
        if d * d > n {
            break;
        }
    }
    divs.push(n);

    for d in 2.. {
        if (n - 1) % d == 0 {
            ddivs.push(d);
            ddivs.push((n - 1) / d);
        }
        if d * d > n - 1 {
            break;
        }
    }
    ddivs.push(n - 1);

    let mut ans = vec![];
    for d in divs {
        let mut N = n;
        while N % d == 0 {
            N /= d;
        }
        if N % d == 1 {
            ans.push(d);
        }
    }

    for d in ddivs {
        if n % d == 1 {
            ans.push(d);
        }
    }

    let uniq: HashSet<usize> = ans.into_iter().collect();

    println!("{}", uniq.len());
}