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 (1.15.1) (judge verdict: Accepted)

use std::io::*;

fn check(x: u64, n: u64) -> u64 {
    let mut m = n / x;
    while m % x == 0 {
        m /= x;
    }
    if m % x == 1 {
        1
    } else {
        0
    }
}

fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let n: u64 = itr.next().unwrap().parse().unwrap();
    let m = n - 1;

    let mut ans = 0;
    let mut i = 1u64;
    // 足すだけ: 1 + x * Kを満たすxがあるか
    while i * i <= m {
        if m % i == 0 {
            ans += 1;
            if m / i != i {
                ans += 1;
            }
        }
        i += 1;
    }
    // 一度でもかけるとKの倍数
    // N mod K == 0 => N + K mod K == 0
    i = 2u64;
    while i * i <= n {
        if n % i == 0 {
            ans += check(i, n);
            if n / i != i {
                ans += check(n / i, n);
            }
        }
        i += 1;
    }
    println!("{}", ans);
}