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;
using ll = long long;

int main() {
	ll N;
	cin>>N;
	ll ans = 0;
	for (ll k = 1;k*k <= N;++k) {
		if ((N-1)%k == 0) {
			if ((N-1)/k != k) {
				ans += 1;
			}
		
			if (k != 1) {
				ans += 1;
			}
		}
	}

	function<void(ll)> compute = [&] (ll x) {
		if (x == 1) return;
		ll z = N;
		while (z%x == 0) {
			z /= x;
		}

		if (z%x == 1) {
			ans += 1;
		}
	};

	for (ll k = 1;k*k <= N;++k) {
		if (N%k == 0) {
			compute(k);
			if (N/k != k) {
				compute(N/k);
			}
		}
	}

	cout<<ans<<endl;
	return 0;
}
