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
/*
@author: sharrad99
*/
#include<bits/stdc++.h>
using namespace std;
int main(){

	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);

	int64_t n;
	cin >> n;

	unordered_set<int64_t> d;
	d.insert(n);
	if(n - 1 > 1) d.insert(n - 1);
	for(int64_t i = 2; i * i <= n; i++) {
		int64_t c = 0;
		int64_t t = n;
		while(t % i == 0) {
			t /= i;
			c++;
		}

		if(t % i == 1) {
			d.insert(i);
			if(c == 0) d.insert((n - 1) / i);
		}
	}

	cout << d.size();
}