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;
typedef long long int ll;
#define rep(i, n) for (int i = 0; i < n; i++)
#define all(x) x.begin(), x.end()

ll divisor(ll a) {
  ll ans = 1;
  for (int i = 2; i <= sqrt(a); i++) {
    int cnt = 0;
    while (a % i == 0) {
      cnt++;
      a /= i;
    }
    ans *= (cnt + 1);
    if (a == 1) break;
  }
  if (a != 1) ans *= 2;
  return ans;
}

int main() {
  ll n;
  cin >> n;
  ll num = divisor(n - 1);
  cout << num + 1 << endl;
}