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: Wrong Answer)

typedef long long ll;
#include <bits/stdc++.h>
using namespace std;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    
    ll n;
    std::cin >> n;
    ll cnt = 0;
    for (ll i = 2; i <= sqrt(n); i++) {
        if(n%i==0){
            set<ll> cand;
            cand.insert(i);
            cand.insert(n/i);
            for (auto e : cand) {
                ll tmp = n;
                while(tmp%e==0)tmp/=e;
                if(tmp%e==1)cnt++;
            }
        }
    }
    cnt++;
    // std::cout << cnt << std::endl;
    // std::cout << n<<" n" << std::endl;
    for (ll i = 2; i <= sqrt(n-1); i++) {
        if((n-1)%i==0){
            cnt++;
            if(i*i!=n-1){
                if((n-1)%((n-1)/i)==0)cnt++;
            }
        }
        // std::cout << i<<" "<<cnt << std::endl;
    }
    cnt++;
    std::cout << cnt << std::endl;
}