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>
#define rep(i,n) for(int i = 0; i < (int)(n); ++i)

using namespace std;
using ll = long long;

int main() {
	ll n, ans = 0;
	cin >> n;
	vector<ll> num;
	rep(k, 2) {
		n -= (ll)k;
		if (n != 1) num.push_back(n);
		for (ll i = 2; i * i <= n; i++) {
			if (n % i == 0) {
				num.push_back(i);
				if (i * i != n) num.push_back(n / i);
			}
		}
		n += k;
	}
	for (ll i : num) {
		ll a = n;
		while(a % i == 0) a /= i;
		if (a % i == 1) {
			ans++;
		}
	}
	cout << ans << endl;
	return 0;
}
