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)

#pragma GCC optimize ("O3")
#pragma GCC optimize ("unroll-loops")
#pragma comment(linker, "/STACK:2000000")

#include "bits/stdc++.h"

using namespace std;

#define pb push_back
#define F first
#define S second
#define f(i,a,b)  for(int i = a; i < b; i++)
// #define endl '\n'
using ll = long long;
using db = long double;
using ii = pair<int, int>;

const int N = 3e5 + 5, LG = 19, MOD = 998244353;
const int SQ =320;
const long double EPS = 1e-7;
void printCase(int cs){
  cout << "Case #" << cs<<": ";
}
vector<ll>getDivs(ll x){
  vector<ll> ret;
  for(ll i = 1; i * i <= x; i++){
    if(x%i==0){
      ret.push_back(i);
      if(i*i!=x)ret.push_back(x/i);
    }
  }
  sort(ret.rbegin(),ret.rend());
  return ret;
}
int32_t main(){
#ifdef ONLINE_JUDGE
    ios_base::sync_with_stdio(0);
    cin.tie(0);
#endif
  ll n;
  cin>> n;
  ll ans = 0;



  auto x = getDivs(n);
  x.pop_back();
  for(auto i : x){
    ll tmp = n;
    while(tmp%i==0)tmp/=i;
    if(tmp%i==1){
      ans++;
    }
  }
  auto y = getDivs(n-1);
  y.pop_back();

  ans+=y.size();

  cout<<ans<<'\n';


  return 0;
}