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 <algorithm>
#include <map>
#include <vector>
#include <cmath>
#define ll long long
using namespace std;

int main(){
	ll n, cnt=0, x, y;
	cin >> n;
	for (ll i=2; i*i<=n; i++){
		if (n%i!=0) continue;

		x = n;
		while(x%i==0) x/=i;
		x%=i;
		if (x==1) cnt++;
		
		y = n, x=n/i;
		while(y%x==0) y/=x;
		y%=x;
		if (y==1) cnt++;

		if (i==n/i && y==1) cnt--;
	}
	n--;
	for(ll i=1; i*i<=n; i++){
		if (n%i==0) {
			cnt++;
			if (i!=n/i) cnt++;
		}
	}
	cout << cnt;
    return 0;
}