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;
constexpr char sp = ' ';

ll N, res;

ll f(ll x, ll p) {
	while (x%p == 0)x /= p;
	return x%p;
}

int main() {
	cin >> N;
	vector<ll>v;
	for (ll i = 1; i*i <= N; i++) {
		if (N%i == 0) {
			v.push_back(i);
			v.push_back(N / i);
		}
		if ((N - 1) % i == 0) {
			v.push_back(i);
			v.push_back((N - 1) / i);
		}
	}
	sort(v.begin(), v.end());
	v.erase(unique(v.begin(), v.end()), v.end());
	for (ll l : v) {
		if (l == 1)continue;
		if (f(N, l) == 1)res++;
	}
	cout << res << endl;
}