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;
using ll = long long;

int main(){
  ll n;
  cin >> n;
  if (n==2){
    cout << 1 << endl;
    return 0;
  }
  int ans = 1;
  n -= 1;
  ll d = 2;
  int m = 0;
  while (n % d == 0){
    n /= d;
    m++;
  }
  ans *= (1 + m);
  d = 3;

  while (n > 1){
    m = 0;
    while (n % d == 0){
      n /= d;
      m++;
    }
    ans *= (1 + m);
    if (d * d > n){
      ans *= 2;
      break;
    }
    d += 2;
  }
  cout << ans + 1 << endl;
  return 0;
}