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)

#![allow(unused, non_snake_case, dead_code, non_upper_case_globals)]
use {
    proconio::{marker::*, *},
    std::{cmp::*, collections::*, mem::*},
};

#[fastout]
fn main() {
    input! {
        n:i64,
    }
    let mut n = n;
    if n == 2 {
        println!("{}", 1);
        return;
    }
    let mut ans = 0;
    for k in 2..=n {
        if k * k > n {
            break;
        }
        // if (n - 1) % k == 0 {
        //     ans += 1;
        // }
        let mut m = n;
        while m % k == 0 {
            m /= k;
        }
        if m % k == 1 {
            ans += 1;
        }
        // dbg!(k, ans);
    }
    // ans += 2;
    println!("{}", ans * 2 + 1);
}