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<algorithm>
#include<vector>
#include<numeric>
#include<string>
#include<cmath>
#include<set>
#include<queue>
#include<deque>
#include<bitset>
#include<iomanip>
#include<cctype>
#include<map>
#include<cstring>

#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define rep2(i,a,b) for (int (i)=a;(i)<(b);(i)++)
#define all(x) (x).begin(),(x).end()

using namespace std;
using ll = long long int;
const int inf = 1001001000;
const long long int Inf = 1001001001001001000;

void print(vector<vector<int>> a){
	for (int i = 0; i < a.size(); i++)
	{
		for (int j=0;j<a[i].size();j++){
			cout << a[i][j] << " ";
		}
		cout << endl;
	}
}

void print(vector<vector<long long int>> a){
	for (int i=0;i<a.size();i++){
		for (int j=0;j<a[i].size();j++){
			cout << a[i][j] << " ";
		}
		cout << endl;
	}
}

void print(vector<int> a){
    int n = a.size();
    for (int j=0;j<n;j++) {
        if (j != n-1) cout << a[j] << " ";
        else cout << a[j] << endl;
    }
}

void print(set<int> a){
	for (auto x:a)
		cout << x << " ";
	cout << endl;
}

//nの約数を返す関数.sortされてない.
vector<long long int> divisors(long long int n){
	vector<long long int> res;
	for (long long int i = 1; i * i <= n;i++){
		if (n % i == 0){
			res.push_back(i);
			if(i * i != n){
				res.push_back(n / i);
			}
		}
	}
	return res;
}


int main()
{
	long long int n;
	cin >> n;
	int ans = 0;

	for (auto x : divisors(n))
	{
		if (x == 1)
			continue;
		long long int temp = n;
		while(temp%x == 0)
			temp /= x;
		temp %= x;
		if (temp == 1)
			ans++;

	}
	ans += divisors(n - 1).size() - 1;
	cout << ans << endl;
	return 0;
}