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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

vector<long long> div(long long n){
  vector<long long> a;
  
  for(int i = 1; i * i <= n; ++i){
    if(n % i != 0) continue;
    a.push_back(i);
    if(i * i != n) a.push_back(n / i);
  }
  
  return a;
}

int main() {
  long long n;
  cin >> n;
  
  long long ans = 0;
  for(long long x : div(n)){
    if(x == 1) continue;
    long long k = n;
    while(k % x == 0) k /= x;
    if(k % x == 1) ans++;
  }
  ans += div(n - 1).size() - 1;
  
  cout << ans << endl;
}