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 ll;

int main() {
	ll N;
	set<ll> s;
	cin >> N;

	for(ll i=1; i*i<=N-1; ++i)
		if((N-1)%i == 0) {
			s.insert(i);
			s.insert((N-1)/i);
		}

	for(ll i=1; i*i<=N; ++i) {
		if(N%i != 0)
			continue;

		ll t=N;
		if(i != 1) {
			while(t%i == 0)
				t /= i;
			if(t%i == 1)
				s.insert(i);
		}

		t=N;
		while(t%(N/i) == 0)
			t /= (N/i);
		if(t%(N/i) == 1)
			s.insert(N/i);
	}

	cout << s.size()-1 << endl;
}