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 <iostream>
#include <vector>
#include <cmath>
#include <iomanip>
#include <algorithm>
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)


int main() {
	long long n,cnt1=1,n2;//n and n-1 is always ans
	cin>>n;
	for(long long i=2;i<=sqrt(n-1);i++){
	    if((n-1)%i==0)cnt1+=2;
	    if(i*i==n-1)cnt1--;
	}
	//cout<<cnt1;//
	cnt1++;
	for(long long i=2;i<=sqrt(n);i++){
	    n2=n;//cout<<n<<endl;
	    while(n2%i==0){
	        n2/=i;
	    }
	    if(n2!=n&&n2%i==1){
	        cnt1++;//cout<<i<<endl;
	    }
	}
	if(n==2)cnt1--;
	cout<<cnt1;
	return 0;
}
