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 SZ(x) (((int)x.size()))
#define all(x) (x).begin(), (x).end()

typedef long long ll;
typedef long double ld;

using namespace std;

ll n, ans, tmp, i;

int main()
{
	cin >> n;
	for (i = 2; i * i <= n; i++)
	{
		tmp = n;
		while (tmp % i == 0)
			tmp /= i;
		if (tmp % i == 1)
			ans++;
	}
	n--;
	for (ll j = 1; j * j <= n; j++)
	{
		if (n % j == 0)
		{
			if (j >= i)
				ans++;
			if (n / j != j && n / j >= i)
				ans++;
		}
	}
	cout << ans + 1;
	return 0;
}
