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>
#define pb push_back
#define mp make_pair
#define ll long long
#define mod 998244353
using namespace std;
ll gcd(ll a, ll b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);

}
int main()
{
    ios_base:: sync_with_stdio(false);
	cin.tie(NULL); cout.tie(NULL);
    ll n;
    cin>>n;
    if(n == 2)
    {
        cout<<1;
        return 0;
    }
    if(n == 1)
    {
        cout<<1;
        return 0;
    }
    ll ans = 2;
    set<ll>s;
    s.insert(n);
    s.insert(n-1);
    //cout<<sqrt(n)<<endl;
    for(ll i=2;i<=sqrt(n);i++)
    {
        ll m = n;
        while(m%i == 0)
            m /= i;
        if((m == 1) || ((m-1)%i == 0))
        {
            //out<<i<<endl;
            if(s.find(i) == s.end())
                ans++;
            s.insert(i);
        }
        ll j=n/i;
        m = n;
        while(m%j == 0)
            m /= j;
        if((m == 1) || ((m-1)%j == 0))
        {
           // cout<<j<<endl;
           if(s.find(j) == s.end())
                ans++;
            s.insert(j);
        }
    }
    cout<<ans;
}