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>
//ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//clock_t start=clock();clock_t end=clock();cout<<(double)(end-start)/CLOCKS_PER_SEC<<endl;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int ui;
typedef pair<int,int> pii;
typedef pair<pii,int> ppii;
typedef pair<int,pii> pipi;
typedef pair<ll,ll> pll;
typedef pair<pll,ll> ppll;
typedef pair<ll,pll> plpl;
typedef tuple<ll,ll,ll> tl;
typedef pair<double,double> pdd;
typedef vector<vector<ll>> mat;
ll mod=1000000007;
ll mod2=998244353;
ll mod3=1000003;
ll mod4=998244853;
ll mod5=1000000009;
ll inf=1000000001;
double pi=3.141592653589793238462643383279L;
double eps=1e-14;
#define rep(i,m,n) for(ll i=m;i<n;i++)
#define rrep(i,n,m) for(ll i=n;i>=m;i--)
#define srep(itr,st) for(auto itr=st.begin();itr!=st.end();itr++)
#define mrep(itr,mp) for(auto itr=mp.begin();itr!=mp.end();itr++)
#define Max(a,b) a=max(a,b)
#define Min(a,b) a=min(a,b)
int dh[4]={1,-1,0,0};
int dw[4]={0,0,1,-1};
int ddh[8]={-1,-1,-1,0,0,1,1,1};
int ddw[8]={-1,0,1,-1,1,-1,0,1};
ll gcd(ll a,ll b){
    if(a<b)swap(a,b);
    if(b==0)return a;
    if(a%b==0)return b;
    return gcd(b,a%b);
}
ll lcm(ll a,ll b){
    ll c=gcd(a,b);
    return a*b/c;
}
ll Pow(ll n,ll k){
    ll ret=1;
    ll now=n;
    while(k>0){
        if(k&1)ret*=now;
        now*=now;
        k/=2;
    }
    return ret;
}
ll beki(ll n,ll k,ll md){
  ll ret=1;
  ll now=n;
    now%=md;
  while(k>0){
    if(k%2==1){
      ret*=now;
      ret%=md;
    }
    now*=now;
    now%=md;
    k/=2;
  }
  return ret;
}
ll gyaku(ll n,ll md){
  return beki(n,md-2,md);
}
ll popcount(ll n){
    ll ret=0;
    ll u=n;
    while(u>0){
        ret+=u%2;
        u/=2;
        
    }
    return ret;
}

int main(){
    ios::sync_with_stdio(false);cin.tie(0);
    ll n;cin>>n;
    ll ans=1;
    unordered_map<ll,ll> mp;
    for(ll i=2;i*i<=n;i++){
        if(n%i==0){
            ll m=n;
            while(m%i==0)m/=i;
            if(m%i==1){
                ans++;
                mp[i]=1;
            }
            if(i*i<n){
                ll u=n/i;
                ll mm=n;
                while(mm%u==0)mm/=u;
                if(mm%u==1){
                    ans++;
                    mp[u]=1;
                }
            }
        }
    }
    n--;
    for(ll i=2;i*i<=n;i++){
        if(n%i==0){
            if(mp[i]==0)ans++;
            if(i*i<n&&mp[n/i]==0)ans++;
        }
    }
    if(n>=2)ans++;
    cout<<ans<<endl;
}