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 int ll;
bool check(ll x,ll y){
	while( y % x == 0){
		y /= x;
	}
	return y % x == 1;
}
void solved(){
	ll n;cin>>n;
	ll ans = 0;
	for(ll i = 1; i * i<= n - 1; i++){
		if((n - 1)%i==0){
			if(i * i == n - 1)ans++;
			else ans += 2;
		}
	}
	ans--;
	for(ll i = 2; i * i <= n; i++){
		if(n % i == 0 && check(i,n)){
			ans++;
		}
	}
	ans++;
	cout<<ans<<endl;
}
int main(){
	solved();
	return 0;
} 