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++ (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
#define il inline
#define re register
#define fo(i,j,k) for(register int i=j;i<=k;i++)
#define inf 0x7fffffff
typedef long long ll;
using namespace std;
il ll read()
{
	re ll x=0,f=1;char c=getchar();
	while(c<'0'||c>'9'){if(c=='-') f=-1;c=getchar();}
	while(c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return f*x;	
}
ll n,ans; 
vector<ll> a;
int main()
{
	cin>>n;
	a.push_back(1);
	ll k=n-1;
	for(re ll i=2;i*i<=k;i++)
	{
		if(k%i==0)
		{
			a.push_back(i);
			if(i*i!=k)
			{
				a.push_back(k/i);
			}
		}
	}
	a.push_back(k);
	for(re ll i=2;i*i<=n;i++)
	{
		for(ll j=n;j>0&&(j%i==0);j/=i)
		{
			ll x=j/i;
			if(x%i==1)
			{
				a.push_back(i);
				break;
			}
		}
	}
	a.push_back(n);
	sort(a.begin(),a.end());
	a.erase(unique(a.begin(),a.end()),a.end());
	ans=a.size()-1;
	printf("%lld\n",ans);
	return 0;
}