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 ll long long

void solve() {
	ll n;
	cin >> n;
	set<ll> s;
	s.insert(2); s.insert(n); s.insert(n - 1);
	for(ll i = 2; i * i <= n - 1; i++) {
		if((n - 1) % i == 0) {
			s.insert(i);
			s.insert((n - 1) / i);
		}
	}
	for(ll i = 2; i * i <= n; i++) {
		if(n % i == 0) {
			ll num = n;
			while(num > 1 && num >= i) {
				if(num % i == 0) num /= i;
				else num = num % i;
			}
			if(num == 1) {
				s.insert(i);
			}
		}
	}
	printf("%d\n", s.size());
}

int main() {
//	freopen("in.txt", "r", stdin);
	solve();
	return 0;
}