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 <iostream>
#include <cstdio>
using namespace std;
typedef long long ll;

ll n, k, x, y, t, ans;
int main() {
	cin >> n;
	if(n==2) return puts("1")*0;
	for(k=1; k*k<=n; k++) {
		if(n%k) continue;
		x = k;
		if(x > 1) {
			for(t=n; t%x==0; ) t/=x;
			if(t%x==1) ans++;
		}
		y = n / k;
		if(y != x) {
			for(t=n; t%y==0; ) t/=y;
			if(t%y==1) ans++;
		}
	}
	for(k=1; k*k<=n-1; k++) {
		if((n-1) % k == 0) {
			if(k==1 || k*k==n-1) ans++;
			else ans += 2;
		}
	}
	cout << ans;
	return 0;
}