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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>

using namespace std;
using ll = long long;

vector<ll> divisor(ll n){
    vector<ll> res;
    for(ll i = 1; i * i <= n; i++){
        if(n % i == 0){
            res.push_back(i);
            if(i != n/i) res.push_back(n/i);
        }
    }
    return res;
}
 
int main(){
  
    ll N;
    cin >> N;
 
    ll ans = 0;
    auto divY = divisor(N-1);
    ans += divY.size() - 1;
    auto divX = divisor(N);
    for(auto&& x : divX){
        if(x == 1) continue;
        ll n = N;
        while(n % x == 0) n /= x;
        if(n % x == 1)
            ans++;
    }
    cout << ans;
  
    return 0;
}