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 <string>
#include <vector>
#include <algorithm>
#include <stdio.h>
#include <sstream>
#include <math.h>
#include <queue>
#include <map>
#include <set>
#include <ios>     // std::left, std::right
#include <iomanip> // std::setw(int), std::setfill(char)
using namespace std;


int main(){
long long N,vm,tmp,tmp_n,ans;
cin>>N;
vector<long long> vn;
std::set<long long> s;
for(long long i=1;i*i<=N;++i){
    if(N%i==0){
        vn.push_back(i);
        vn.push_back(N/i);
    }
    if((N-1)%i==0){
        s.insert(i);
        s.insert((N-1)/i);
    }
}
sort(vn.begin(), vn.end());

//Kで一回は割れるときcheck

for(long long i=1;i<vn.size();++i){
    tmp=vn[i];
    tmp_n=N;
    while(tmp_n%tmp==0){
       tmp_n/=tmp; 
    }
    if((tmp_n-1)%tmp==0){
        s.insert(tmp);

    }
}

tmp=s.size();
ans=tmp-1;

cout<<ans<<"\n";

return 0;
}