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)

macro_rules ! input { ( source = $ s : expr , $ ( $ r : tt ) * ) => { let mut iter = $ s . split_whitespace ( ) ; input_inner ! { iter , $ ( $ r ) * } } ; ( iter = $ iter : ident , $ ( $ r : tt ) * ) => { let s = { use std :: io :: Read ; let mut s = String :: new ( ) ; std :: io :: stdin ( ) . read_to_string ( & mut s ) . unwrap ( ) ; s } ; let mut $ iter = s . split_whitespace ( ) ; input_inner ! { $ iter , $ ( $ r ) * } } ; ( $ ( $ r : tt ) * ) => { let s = { use std :: io :: Read ; let mut s = String :: new ( ) ; std :: io :: stdin ( ) . read_to_string ( & mut s ) . unwrap ( ) ; s } ; let mut iter = s . split_whitespace ( ) ; input_inner ! { iter , $ ( $ r ) * } } ; }
macro_rules ! input_inner { ( $ iter : expr ) => { } ; ( $ iter : expr , ) => { } ; ( $ iter : expr , mut $ var : ident : $ t : tt $ ( $ r : tt ) * ) => { let mut $ var = read_value ! ( $ iter , $ t ) ; input_inner ! { $ iter $ ( $ r ) * } } ; ( $ iter : expr , mut $ var : ident $ ( $ r : tt ) * ) => { input_inner ! { $ iter , mut $ var : usize $ ( $ r ) * } } ; ( $ iter : expr , $ var : ident : $ t : tt $ ( $ r : tt ) * ) => { let $ var = read_value ! ( $ iter , $ t ) ; input_inner ! { $ iter $ ( $ r ) * } } ; ( $ iter : expr , $ var : ident $ ( $ r : tt ) * ) => { input_inner ! { $ iter , $ var : usize $ ( $ r ) * } } ; }
macro_rules ! read_value { ( $ iter : expr , ( $ ( $ t : tt ) ,* ) ) => { ( $ ( read_value ! ( $ iter , $ t ) ) ,* ) } ; ( $ iter : expr , [ $ t : tt ; $ len : expr ] ) => { ( 0 ..$ len ) . map ( | _ | read_value ! ( $ iter , $ t ) ) . collect ::< Vec < _ >> ( ) } ; ( $ iter : expr , { chars : $ base : expr } ) => { read_value ! ( $ iter , String ) . chars ( ) . map ( | c | ( c as u8 - $ base as u8 ) as usize ) . collect ::< Vec < usize >> ( ) } ; ( $ iter : expr , { char : $ base : expr } ) => { read_value ! ( $ iter , { chars : $ base } ) [ 0 ] } ; ( $ iter : expr , chars ) => { read_value ! ( $ iter , String ) . chars ( ) . collect ::< Vec < char >> ( ) } ; ( $ iter : expr , char ) => { read_value ! ( $ iter , chars ) [ 0 ] } ; ( $ iter : expr , usize1 ) => { read_value ! ( $ iter , usize ) - 1 } ; ( $ iter : expr , $ t : ty ) => { $ iter . next ( ) . unwrap ( ) . parse ::<$ t > ( ) . unwrap ( ) } ; }
pub fn prime_factors(n: usize) -> HashMap<usize, usize> {
    let mut factors = HashMap::new();
    let mut n = n;
    for i in 2..(n as f32).sqrt() as usize + 1 {
        while n % i == 0 {
            n /= i;
            *factors.entry(i).or_insert(0) += 1;
        }
    }
    if n > 1 {
        *factors.entry(n).or_insert(0) += 1;
    }
    factors
}
pub fn gcd(a: usize, b: usize) -> usize {
    let mut a = a;
    let mut b = b;
    loop {
        if b == 0 {
            return a;
        }
        a %= b;
        if a == 0 {
            return b;
        }
        b %= a;
    }
}
use std::collections::HashMap;
fn main() {
    input! { n };
    let fa1 = prime_factors(n - 1);
    let mut y = 1;
    for &v in fa1.values() {
        y *= v + 1;
    }
    // y -= 1;
    for i in 2..(n as f32).sqrt() as usize + 1 {
        if n % i == 0 {
            {
                let j = i;
                let mut m = n;
                while m > 0 && m % j == 0 {
                    m /= j;
                }
                if m % j == 1 {
                    y += 1;
                }
            }
            if i * i != n {
                let j = n / i;
                let mut m = n;
                while m > 0 && m % j == 0 {
                    m /= j;
                }
                if m % j == 1 {
                    y += 1;
                }
            }
        }
    }
    let ans = y;
    println!("{}", ans);
}