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 <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <string>
#include <queue>
 
using namespace std;
 
int main()
{
	long long n;
	cin>>n;
	long long cnt=1;
	if(n>2)cnt++;
	for(long long i=2;i*i<n;i++){
		if((n-1)%i==0){
			cnt++;
			if((n-1)/i!=i)
				cnt++;
		}
	}
	for(long long i=2;i*i<n+1;i++){
		if(n%i==0){
			long long cur=n;
			while(cur%i==0)cur/=i;
			if(cur%i==1)cnt++;
		}
	}
	cout<<cnt<<endl;
	return 0;
}