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 int long long 

main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	int N; cin >> N;
	unordered_set<int> S;
	//all factors of N-1
	for (int i=1; i*i<=N-1; i++){
		if ((N-1)%i==0){
			S.insert(i);
			S.insert((N-1)/i);
		}
	}
	for (int i=2; i*i<=N; i++){
		if (N%i!=0) continue;
		int k = i;
		while (k<=N){
			if (N/k % i==1){
				S.insert(k);
				break;
			}
			k *= i;
		}
		k = N/i;
	}
	S.insert(N);
	S.erase(1);
	cout << (int) S.size();
}