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;

long long n;
set<long long> se;

int main() {
	scanf("%lld", &n);
	se.insert(1), se.insert(n);
	for(int i = 2; 1LL * i * i <= n; ++i) {
		if(n % i) continue;
		long long d = i, temp = n / d;
		while(temp % d == 0) temp /= d;
		if(temp % d == 1) se.insert(d);
		d = n / i, temp = i;
		while(temp % d == 0) temp /= d;
		if(temp % d == 1) se.insert(d);
	}
	--n;
	for(int i = 1; 1LL * i * i <= n; ++i) {
		if(n % i) continue;
		se.insert(i), se.insert(n / i);
	} se.erase(1);
	printf("%d\n", int(se.size()));
	return 0;
}