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;
using ll = long long;

int main() {
    ll N; cin >> N;
    ll nt = N-1;
    ll ans = 1;

    map<int, int> m;
    for (int i = 2; i <= sqrt(nt); i++)
    {
        while(nt%i==0){
            nt/=i;
            m[i]++;
        }
    }
    if(nt>1) m[nt]++;

    for(auto i : m){
        //cout << i.first << " " << i.second << endl;
        ans *= i.second+1;
    }
    ans--; // remove 1
    //cout << ans << endl;

    ans++; //N
    for (int i = 2; i <= sqrt(N); i++)
    {
        ll nt = N;
        if(nt%i!=0) continue;
        while(nt%i==0) nt /= i;
        if(nt%i==1) ans++;
    }
    cout << ans << endl;
}