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;
use std::collections::BTreeSet;

// n mod k = 1 なら k で可
// n = mk^e (m⊥k) のとき、m mod k = 1 なら k で可

fn main() {
    input!(n: u64);
    let mut s = BTreeSet::new();
    s.extend(divisors(n - 1));
    for k in divisors(n) {
        let mut m = n;
        while m % k == 0 {
            m /= k;
        }
        if m % k == 1 {
            s.insert(k);
        }
    }
    println!("{}", s.len())
}

fn divisors(n: u64) -> Vec<u64> {
    let mut res = vec![n];
    for d in 2.. {
        if d * d > n {
            break;
        }
        if n % d == 0 {
            res.push(d);
            res.push(n / d)
        }
    }
    res
}