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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); i++)
using namespace std;
using P = pair<int, int>;
typedef long long ll;

//約数列挙 計算量O(sqrt(n))
vector<long long> divisor(long long n) {
    vector<long long> ret;
    for (long long i = 1; i * i <= n; i++) {
        if (n % i == 0) {
            ret.push_back(i);
            if (i * i != n) ret.push_back(n / i);
        }
    }
    sort(ret.begin(), ret.end()); // 昇順に並べる
    return ret;
}

int main() {
  ll n;
  cin >> n;
  map<ll, ll> mp;
  vector<ll> ans1 = divisor(n - 1);
  vector<ll> ans2 = divisor(n);
  for (int i = 1; i < ans1.size(); i++){
    mp[ans1[i]]++;
  }
  mp[n]++;
  for (int i = 1; i < ans2.size(); i++){
    ll num = n;
    while(num % ans2[i] == 0){
      num /= ans2[i];
    }
    if(num % ans2[i] == 1) mp[ans2[i]]++;
  }
  cout << mp.size() << endl;
}