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 <cstdio>
#include <algorithm>
#include <set>
using namespace std;
int main() {
	long long N, ans = 0;
	scanf("%lld", &N);
	set<long long> guess;
	for (long long i = 1; i * i <= N; ++i) {
		if (N % i == 0) guess.insert(i), guess.insert(N / i);
	}
	for (long long i = 1; i * i < N; ++i) {
		if ((N - 1) % i == 0) guess.insert(i), guess.insert((N - 1) / i);
	}
	guess.erase(1);
	for (long long i: guess) {
		long long T = N;
		while (T % i == 0) T /= i;
		if ((T - 1) % i == 0) ++ans;
	}
	printf("%lld", ans);
	return 0;
}