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)

/// kazuki08
/*
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
*/
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
#define int long long
typedef long long ll;
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>ordered_set;
typedef long double ld;
typedef vector<int> vi;
typedef pair<int,int> pi;
typedef vector<pi> vpi;
#define ar array
//#define inf 1000000000
#define mod 1000000007
#define dmp(x) cerr<<"line "<<__LINE__<<" "<<#x<<":"<<x<<endl
#define fs first
#define sc second
#define pb push_back
#define all(c) (c).begin(),(c).end()
#define mt make_tuple
#define IOS ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
template<class S,class T> ostream& operator<<(ostream& o,const pair<S,T> &p){
    return o<<"("<<p.fs<<","<<p.sc<<")";
}
template<class T> ostream& operator<<(ostream& o,const vector<T> &vc){
    o<<"{";
    for(const T& v:vc) o<<v<<",";
    o<<"}";
    return o;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int n,ans=0;

bool chk(int x){
  int cur = n;
  while(cur%x==0){
    cur/=x;
  }
  return cur%x==1;
}

signed main(){
IOS;
//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);
  
  cin >> n;
  int cur = n-1;
  vi v;
  for(int i=2;i*i<=cur;i++){
    if(cur%i==0){
      v.pb(i);
      if(i*i!=cur){
        v.pb(cur/i);
      }
    }
  }
  if(cur>1)
  v.pb(cur);
  ans+=v.size();v.clear();
  // cout << v.size() << endl;
  cur = n;
  for(int i=2;i*i<=cur;i++){
    if(cur%i==0){
      v.pb(i);
      if(i*i!=cur){
        v.pb(cur/i);
      }
    }
  }
  v.pb(cur);
  for(auto t:v){
    if(chk(t))ans++;
  }
  cout << ans << endl;

return 0;
}
///....