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
// F - Division or Substraction
#include <bits/stdc++.h>
using namespace std;
using ll = long long;

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

int main(){
	ll N; cin>>N;
	ll ans = divisors(N-1).size() - 1;
	for(ll k:divisors(N)){
		if(k == 1) continue;
		ll n = N;
		while(n%k == 0) n /= k;
		if(n%k == 1) ++ans;
	}
	cout<< ans <<endl;
}