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;

#define ll long long

ll n;
set<ll> st;

vector<ll> divisor(ll n){
	vector<ll> res;
	for (ll i=1;i*i<=n;i++){
		if (n%i==0){
			res.push_back(i);
			if (n/i!=i) res.push_back(n/i);
		}
	}
	return res;
}

void solve(ll n,ll m){
	vector<ll> v=divisor(n);
	for (ll i=0;i<v.size();i++){
		ll k=v[i];
		if (k==1) continue;
		ll d=m;
		while (d>=k){
			if (d%k==0) d/=k;
			else break;
		}
		d%=k;
		if (d==1) st.insert(k);
	}
}

int main(){
	cin>>n;
	solve(n,n);
	solve(n-1,n);
	cout<<st.size()<<endl;
	return 0;
}