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.

Examples

Input

6


Output

3


Input

3141


Output

13


Input

314159265358


Output

9
#include<bits/stdc++.h>
using namespace std;
long long N,M;
bool check(long long x){
	if(x==1)return 0; 
	long long tmp=M;
	while(tmp%x==0)tmp/=x;
	if(tmp%x==1)return 1;
	return 0;
}
bool check2(){
	long long tmp=sqrt(N);
	tmp=tmp*tmp;
	return tmp==N;
}
int main(){
	cin>>N;M=N;
	int cnt=0;
	for(long long i=1;i<sqrt(N);i++)
		if(N%i==0)
			cnt+=check(i),cnt+=check(N/i);
	if(check2())cnt+=check(sqrt(N));
	N--;
	for(long long i=1;i<sqrt(N);i++)
		if(N%i==0)
			cnt+=check(i),cnt+=check(N/i);
	if(check2())cnt+=check(sqrt(N));
	cout<<cnt; 
}