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)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
struct edge{ll to,cost;};
const int inf = 1<<27;
const ll INF = 1LL<<60;
const int COMBMAX = 1001001;
const ll MOD = 1000000007;
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define eachdo(v, e) for(const auto& e : (v))
#define all(v) (v).begin(), (v).end()
#define lower_index(v, e) (ll)distance((v).begin(), lower_bound((v).begin(), (v).end(), e))
#define upper_index(v, e) (ll)distance((v).begin(), upper_bound((v).begin(), (v).end(), e))
ll mpow(ll a, ll n, ll mod = MOD){ll res = 1; while(n > 0){if(n & 1)res = res * a % mod; a = a * a % mod; n >>= 1;} return res;}
void YesNo(bool j){cout << (j ? "Yes" : "No") << endl; return;}
void yesno(bool j){cout << (j ? "yes" : "no") << endl; return;}
template<class Head> void pt(Head&& head){cout << head << endl; return;}
template<class Head, class... Tail> void pt(Head&& head, Tail&&... tail){cout << head << " "; pt(forward<Tail>(tail)...);}
template<class T> void debug(T v){rep(i, v.size()) cout << v[i] << " " ; cout << endl;}
template<class T> void debug2(T v){rep(i, v.size()){rep(j, v[i].size()) cout << v[i][j] << " " ; cout << endl;}}
template<class T1, class T2> ll bcount(T1 v, T2 a){return upper_index(v, a) - lower_index(v, a);} 
template<class T1, class T2> inline bool chmin(T1 &a, T2 b){if(a > b){a = b; return true;} return false;}
template<class T1, class T2> inline bool chmax(T1 &a, T2 b){if(a < b){a = b; return true;} return false;}
template<class T1, class T2> void sortTwoVecter(vector<T1> &f, vector<T2> &s, bool isReverse = false){
    vector<pair<T1, T2>> p;
    rep(i, f.size()) p.emplace_back(make_pair(f[i], s[i]));
    if (!isReverse) sort(p.begin(), p.end());
    else sort(p.rbegin(), p.rend());
    rep(i, f.size()) tie(f[i], s[i]) = p[i];
    return;
}

ll f(ll n, ll m){
    while(n % m == 0) n /= m;
    return n;
}

int main(){
    ll N; cin >> N;
    ll ans = 2;
    for(ll i = 2; i <= pow(N, 0.5); i++){
        if ((N - 1) % i == 0){
            ans++;
            if (i * i == N - 1) continue;
            ans++;
        }else if (N % i == 0){
            if (f(N, i) % i == 1) ans++;
            if (i * i == N) continue;
            if (f(N, (N / i)) % (N / i) == 1) ans++;
        }
    }
    cout << ans << endl;
    return 0;
}