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>
#define f first
#define s second
#define pb push_back
#define M (L+R)/2
#define N 500009
#define inf 99999999999
#define ll long long
#define mod 1000000007
#define sz size();
using namespace std;
ll n,ans,x,y;
int main(){

	ios::sync_with_stdio(0);
	
	cin>>n;
	x=n-1;

	for(int i=1;i<=sqrt(x);i++){
		if((x%i)==0){
			if(i==sqrt(x))ans++;
			else ans+=2;
		}
	}

	for(int i=2;i<=sqrt(n);i++){
		if(n%i==0){
			x=n;
			while(x%i==0){
				x/=i;
			}
			if((x>i && x%i==1) || x==1){
				ans++;
			}
		}
	}
	cout<<ans<<endl;
	return 0;
}

