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 <cstdio>
#include<iostream>
using namespace std;
typedef long long ll;
int check(ll n, ll div){
     if(n % div==0)
        return check(n/div, div);
    else  return (n % div == 1) ;
}

int main(){
	ll n;
	int res = 0;

	scanf("%lld", &n);
	for(ll i=2; i*i <= n; i++) {
		if(check(n, i)) {
			res++;
			if(i != n / i && check(n, n/i))
				res++;
		}
	}
	if(n==2)     printf("%d",res+1);
	else      printf("%d",  res+2);

	return 0;
}
