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 <iostream>
#include <cmath>

typedef long long ll;

using namespace std;

int main() {
	ll n;
	cin >> n;
	
	int ans = 0;
	for(int i = 1; i * 1ll * i <= n - 1; i++) {
		if((n - 1) % i == 0) {
			if(i > 1) ans++;
			if((n - 1) / i != i) ans++;
		}
	}
	
	auto check = [&](ll v) {
		ll x = n;
		while(x % v == 0) x /= v;
		if(x % v == 1) ans++;
	};
	
	for(int i = 1; i * 1ll * i <= n; i++) {
		if(n % i == 0) {
			if(i > 1) check(i);
			if((n / i) != i) check(n / i);
		}
	}
	
	cout << ans << endl;
}