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;
#define ALL(a) (a).begin(),(a).end()
#define rALL(a) (a).rbegin(),(a).rend()
typedef pair<int, int> Pint;
typedef pair<int64_t, int64_t> Pll;
typedef int64_t ll;

map< int64_t, int > prime_factor(int64_t n) {
  map< int64_t, int > ret;
  for(int64_t i = 2; i * i <= n; i++) {
    while(n % i == 0) {
      ret[i]++;
      n /= i;
    }
  }
  if(n != 1) ret[n] = 1;
  return ret;
}


int main() {
    int64_t N, ans = 1, sum = 1;
    cin >> N;

    for (int64_t i = 2; i * i <= N + 100; i++){
        break;
        if (i == N){
            break;
        }
        if (N % i == 0){
            int64_t a = N;
            while(a % i == 0){
                a /= i;
            }
            if (a % i == 1){
                ans++;
            }
        }
    }
    auto T = prime_factor(N - 1);
    for (auto y: T){
        sum *= y.second + 1;
    }
    ans += sum - 1;
    cout << ans + 1 << endl;
}