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)

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//**************************************Accepted*****************************************//

#include<bits/stdc++.h>

#define all(x) x.begin(),x.end()
#define pb push_back
#define mp make_pair
#define Unique(x) x.erase(unique(all(x)),x.end());

#define CIN_FILE "input.txt"
#define COUT_FILE "output.txt"

#define Pig (OO)

using namespace std;

const int dx[]={-1,0,1,0};
const int dy[]={0,-1,0,1};

const int Days[]={-1,31,28,31,30,31,30,31,31,30,31,30,31};

inline int read()
{
	#define gc getchar
	int ans=0;
	bool f=1;
	char ch=gc();
	
	while(!isdigit(ch))f^=ch=='-',ch=gc();
	
	while(isdigit(ch))ans=((ans<<2)+ans<<1)+(ch^48),ch=gc();
	
	return f?ans:-ans;
	
	#undef gc
}

typedef long long ll;

ll n;
map<ll,bool> pr;

bool isp(ll x)
{
	if(pr[x]==1)return 1;
	for(ll i=2;i*i<=x;i++)if(x%i==0)return false;
	pr[x]=1;return true;
}

ll cnt=1ll;

void DFS(ll x,ll s)
{
	if(isp(x)){cnt*=2;return;}
	
	ll t=x;
	ll ccnt=1;
	
	while(t%s==0&&t>1)
	{
		t/=s;
		
		ccnt++;
	}
	cnt*=ccnt;
	
	for(ll i=s+1;;i++){DFS(t,i);return;}
}

ll m;

int main(int argc, char const *argv[])
{
	ios::sync_with_stdio(false);
	
	cin>>n;n--;
	
	DFS(n,2);
	
	cout<<cnt+1ll;
	
	//printf("Time used = %.12f",(double)(clock())/CLOCKS_PER_SEC);
	return 0;
}

/*  GOD BLESS  */