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<bits/stdc++.h>
using namespace std;

// vector<bool> prime(1000002,true);
typedef long long int ll;

// void sieve(){
// 	for(ll i = 2;i<= 1000000;i++){
// 		if(prime[i]){
// 			for(ll j = i * i;j <= 1000000;j+=i)
// 				prime[j] = false;
// 		}
// 	}
// }


ll no_div(ll n){
	ll cnt = 0;
	for(ll i =1;i*i <= n;i++){
		if(n % i == 0){
			if(n / i == i)
				cnt++; // count only once
			else
				cnt+=2; //count both
		}
	}

	return cnt - 1;
}

int main(){
	ll temp;
	ll n;
	cin>>n;
	// prime[0] = false;
	// prime[1] = false;
	// sieve();
	ll x = no_div(n-1);
	//cout<<x<<endl;
	for(ll i = 2;i * i<=n;i++){
		if(n%i == 0){
			temp = n;
			while(temp % i == 0)
				temp/=i;

			if(temp % i == 1)
				x++;
		}
	}

	cout<<x+1<<"\n";

	return 0;
}