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;
typedef long long LL;
int main(){
	LL n,n1;scanf("%lld",&n);
	if(n==2)return puts("1"),0;
	int q=sqrt(n),ans=0;
	for(int i=2;i<=q;i++){
		if(n%i==0){
			LL w=n/i;n1=n;
			while(n1%i==0)n1/=i;
			if(n1%i==1)ans++;
			n1=n;
			if(w==i)continue;
			while(n1%w==0)n1/=w;
			if(n1%w==1)ans++;
		}
		else if((n-1)%i==0){
			ans+=2;
			if(1ll*i*i==n-1)ans--;
		}
	}
	printf("%d",ans+2);
	return 0;
}