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;

int main() {
	long long n, i, ans = 1;
	cin >> n;
	for (i = 1; i * i <= n-1;i++){
		if((n-1)%i==0){
			if(i>1)
				ans++;
			if(i*i<n-1&&n-1>1)
				ans++;
		}
	}
	for (i = 2; i * i <= n;i++){
		if(n%i==0){
			long long cn = n;
			while(cn%i==0)
				cn /= i;
			if(cn%i==1)
				ans++;
			if(i*i<n){
				cn = n;
				while (cn % (n/i) == 0)
					cn /= (n/i);
				if (cn % (n/i) == 1)
					ans++;
			}
		}
	}
	cout << ans << endl;
}