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: Accepted)

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
#define endl '\n'
#define pi pair<int, int>
#define f first
#define s second

long long n;
vector<int> v;

int main(){
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	
	cin >> n;
	
	long long ret = 1, x = n - 1;
	for(long long i = 2; i * i <= n; i++){
		int c;
		for(c = 1; x % i == 0; x /= i) c++;
		ret *= c;
		
		if(n % i == 0){
			v.push_back(i);
			if(i * i != n) v.push_back(n / i);
		}
	}
	ret *= (x > 1) + 1;
	
	for(long long i : v){
		x = n;
		while(x % i == 0) x /= i;
		ret += x % i == 1;
	}
	
	cout << ret << endl;

	return 0;
}