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;

int fp;
long long n;

vector<long long> g(long long a) {
	vector<long long> dr;
	for(long long i = 2; i * i <= a; ++i) {
		if(a % i == 0) {
			dr.push_back(i);
			if(a / i != i) dr.push_back(a / i);
		}
	}
	if(a > 1) dr.push_back(a);
	return dr;
}

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin >> n;
	fp = g(n - 1).size();
	vector<long long> v = g(n);
	for(long long i : v) {
		long long p = n;
		while(p % i == 0) {
			p /= i;
		}
		if(p % i == 1) ++fp;
	}
	cout << fp << endl;
	return 0;
}
