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;
using int64 = long long;

vector<int64> divisor(int64 n){
	vector<int64> res;
	for(int64 i = 1; i * i <= n; i++){
	if(n % i == 0){
			res.push_back(i);
			if(i != n / i) res.push_back(n / i);
		}
	}
	return res;
}

int main(){
	int64 n;
	cin >> n;
	vector<int64> d = divisor(n - 1);
	int ans = d.size() + 1;
	cout << ans << endl;
}