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 ll long long
#define all(V) V.begin(),V.end()
#define MOD 1000000007

int main() {
	ll N, x, y;
	cin >> N;
	if (N == 2) {
		cout << 1 << endl;
		return 0;
	}
	ll cnt = 2;

	for (ll i = 2;i * i <= N;i++) {
		x = N;
		if (x % i)continue;
		while (x % i == 0) {
			x = x / i;
		}
		if ((x - 1) % i == 0) {
			cnt++;
		}
	}
	N--;
	for (ll i = 2;i * i <= N;i++) {
		if (N % i == 0) {
			if (i * i == N)cnt++;
			else cnt += 2;
		}
	}
	cout << cnt << endl;
}