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
#pragma GCC optimize("Ofast", "unroll-loops")

#include <bits/stdc++.h>

using namespace std;

#define ll long long

ll last(ll N, ll K) {
	while (N % K == 0 && N >= K)
		N /= K;
	return N % K;
}

int main() {
	ll N; cin >> N;
	int res = 1; // K = N : ok
	if (N > 2) ++res; // K = N - 1 : ok
	ll k;
	for (k = 2; k * k <= N; ++k)
		if (last(N, k) == 1)
			++res;
	for (ll i = 2; i * i <= N - 1; ++i) {
		if ((N - 1) % i == 0) {
			if (i >= k) ++res;
			if (i * i != N - 1 && (N - 1) / i >= k)
				++res;
		}
	}
	cout << res << endl;
	return 0;
}