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<bits/stdc++.h>
using namespace std;
#define FOR(i,l,r) for(size_t i=(l);i<(r);++i)
#define REP(i,n) FOR(i,0,n)
#define int long long

int kosuu(int x){
    map<int,int>memo;
    for(int i=2;i<=sqrt(x);i++){
        while(x%i==0){
            x /=i;
            memo[i]++;
        }
    }
    int ans=1;
    for(auto it=memo.begin();it!=memo.end();it++){
        ans *=(it->second+1);
    }
    if(x!=1){
        ans*=2;
    }
    return ans;
}

signed main(){
    int N;
    cin >>N;
    int ans=kosuu(N-1);
    for(int i=2;i<=sqrt(N);i++){
        int N1=N;
      	if(N1%i!=0){
          continue;
        }
      	while(N1%i==0){
            N1 /=i;
        }
        if(N1%i==1){
            ans++;
        }
        int N2=N;
        int j=N/i;
        if(i==j){
            continue;
        }
        while(N2%j==0){
            N2 /=j;
        }
        if(N2%j==1){
            ans++;
        }
    }
    cout<<ans<<endl;
}