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::{input, fastout};

#[allow(unused_imports)]
use std::collections::*;
#[allow(unused_imports)]
use std::cmp::*;

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

    fn calc(n: usize, k: usize) -> usize {
        if n % k == 0 {
            calc(n / k, k)
        }
        else {
            n
        }
    }

    let mut ans = 1;

    let mut k = 2;

    while k * k <= n {
        if n % k == 0 {
            if calc(n, k) % k == 1 { ans += 1; }
            let rev = n / k;
            if rev != k {
                if calc(n, rev) % rev == 1 { ans += 1; }
            }
        };
        k += 1;
    }

    let m = n - 1;
    k = 2;

    ans += 1;
    while k * k <= m {
        if m % k == 0 {
            if n % k == 1 { ans += 1; }
            let rev = m / k;
            if rev != k {
                if n % rev == 1 { ans += 1; }
            }
        };
        k += 1;
    }
    
    println!("{}", ans);
}