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;
typedef long long ll;

const ll maxs = 1e6+5;
const ll lmaxs = 22;
ll mod = 1e9+7;
ll oo = 1e15;

#define IOS ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define I insert
#define endl '\n'


void pre(){

}

vector<ll> div(ll n){
    vector<ll> res;
    for(ll i=1;i*i<=n;i++){
        if(n%i==0){
            res.pb(i);
            if(i != n/i)
                res.pb(n/i);
        }
    }
    sort(res.begin(),res.end());
    return res;
}

void solve(){
    ll n;
    cin>>n;
    ll ans=0;
    vector<ll> DIV_N = div(n);
    for(auto k : DIV_N){
        if(k==1)
            continue;
        ll m = n;
        while(m%k==0) m/=k;
        if(m%k==1)ans++;
    }
    vector<ll> DIV_N_1 = div(n-1);
    ans += DIV_N_1.size();
    ans--; // for 1;
    cout << ans;
}

int main(){
    
    IOS;
    pre();
    ll T = 1;
    //cin>>T;
    while(T--){
        solve();   
    }
    return 0;
}