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: Wrong Answer)

/*** author: yuji9511 ***/
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using lpair = pair<ll, ll>;
const ll MOD = 1e9+7;
const ll INF = 1e18;
#define rep(i,m,n) for(ll i=(m);i<(n);i++)
#define rrep(i,m,n) for(ll i=(m);i>=(n);i--)
#define printa(x,n) for(ll i=0;i<n;i++){cout<<(x[i])<<" \n"[i==n-1];};
void print() {}
template <class H,class... T>
void print(H&& h, T&&... t){cout<<h<<" \n"[sizeof...(t)==0];print(forward<T>(t)...);}
struct Integer{
public:
    ll gcd(ll a,ll b){return b ? gcd(b, a % b) : a;}
    ll lcm(ll a,ll b){return a / gcd(a, b) * b;}

    bool isPrime(ll x){
        if(x < 2) return false;
        if(x == 2) return true;
        if(x % 2 == 0) return false;
        for(ll i = 3;i <= sqrt(x) + 1;i += 2) if(x % i == 0) return false;
        return true;
    }

    vector<ll> divisor(ll M){
        vector<ll> dd;
        for(ll i = 1; i * i <= M; i++){
            if(M % i == 0){
                dd.push_back(i);
                if(i * i != M) dd.push_back(M / i);
            }
        }
        sort(dd.begin(), dd.end());
        return dd;
    }

    vector<ll> factor(ll M){ //素因数分解
        vector<ll> dd;
        if(M == 1){
            dd.push_back(1);
            return dd;
        }
        for(ll i = 2; i*i <= M; i++){
            while(M % i == 0){
                dd.push_back(i);
                M /= i;
            }
        }
        if(M != 1) dd.push_back(M);
        sort(dd.begin(), dd.end());
        return dd;
    }
};

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll N;
    cin >> N;
    Integer it;
    vector<ll> dd = it.divisor(N);
    vector<ll> dd2 = it.divisor(N-1);
    map<ll,ll> mp;
    for(auto &e: dd2){
        if(e != 1) mp[e]++;
    }
    for(auto &e: dd){
        if(e == 1) continue;
        ll v = N / e;
        if(v % e == 1) mp[e]++;
    }
    mp[N]++;
    ll ans = mp.size();
    print(ans);
    // printa(dd, dd.size());
    // printa(dd2, dd2.size());



    

}