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;

#define x first
#define y second
#define mp make_pair
#define pb push_back

typedef long long ll;

const ll MOD = 1e9+7;
const ll INF = 1e9+5;

int main() {
	ll n;
	scanf("%lld", &n);
	ll cnt = 0;
	for (ll k=2 ; k<=min(n, 1000000LL) ; k++) {
		ll m = n;
		while (m%k == 0) m/=k;
		if (m%k == 1) cnt++;
	}
	if (n > 1000000LL) cnt++;
	n--;
	for (ll i=1 ; i*i<=n ; i++) {
		if (n%i == 0) {
			if (n/i > 1000000LL) cnt++;
		}
	}
	printf("%lld\n", cnt);
	return 0;
}