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;
typedef long long int ll;
ll n;
set<ll>st;
inline ll read()
{
    register ll num=0,neg=1;
    register char ch=getchar();
    while(!isdigit(ch)&&ch!='-')
    {
        ch=getchar();
    }
    if(ch=='-')
    {
        neg=-1;
        ch=getchar();
    }
    while(isdigit(ch))
    {
        num=(num<<3)+(num<<1)+(ch-'0');
        ch=getchar();
    }
    return num*neg;
} 
inline void check(ll kk)
{
	ll cur=n;
	while(cur%kk==0)
	{
		cur/=kk;	
	} 
	cur%=kk;
	if(cur==1)
	{
		st.insert(kk);
	}
}
int main()
{
	n=read();
	if(n==2)
	{
		return puts("1"),0;
	}
	for(register int i=1;i<=sqrt(n-1);i++)
	{
		if((n-1)%i==0)
		{
			if(i!=1)
			{
				st.insert(i);
			}
			st.insert((n-1)/i);
		}
	}
	for(register int i=1;i<=sqrt(n);i++)
	{
		if(n%i==0)
		{
			if(i!=1)
			{
				check(i);
			}
			check(n/i);
		}
	}
	printf("%d\n",st.size());
}