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 <algorithm>
using namespace std;
typedef long long ll;


int main(void) {
	ll n, a, i, b, ans = 2;

	cin >> n;

	if (n == 2) {
		cout << 1 << endl;
		return 0;
	}

	for (a = 2; a * a <= n; a++) {
		if (n % a == 0) {
			b = n;
			while (b % a == 0) b /= a;
			if (b % a == 1) ans++;

			if (a * a == n) continue;
			b = n;
			while (b % (n / a) == 0) b /= (n / a);
			if (b % (n / a) == 1) ans++;
		}
		if (n % a == 1) ans += 2;
		if (a * a == n - 1) ans--;
	}

	cout << ans << endl;

	return 0;
}