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;
typedef long int li;
typedef long double lb;
#define rep(i,j,n) for (ll i = j; i < (n); i++)
#define repr(i,j,n) for(ll i = j; i >= (n); i--)
#define all(x) (x).begin(),(x).end()
#define CLR(mat,f) memset(mat, f, sizeof(mat))
#define IN(a, b, x) (a<=x&&x<b)
#define out(ans) cout << ans << endl
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
typedef pair<ll,ll>P;
const ll mod=1e9+7;
const ll INF = 1LL<<60;
const ll m=1LL<<32;



int main(){

  ll n;cin>>n;

  set<ll>ans;
  for(ll i=2;i*i<=n-1;i++){
    if((n-1)%i==0)ans.insert(i),ans.insert((n-1)/i);
  }

  ans.insert(n-1);
  set<ll>se;
  for(ll i=2;i*i<=n;i++){
    if(n%i==0)se.insert(i),se.insert(n/i);
  }
  se.insert(n);

  for(auto s:se){
    ll k=n;
    while(k%s==0){
      k/=s;
    }
    if(k%s==1)ans.insert(s);
  }

  out(ans.size()+(n==2?-1:0));




  return 0;
}