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;
#define rep(i,n) for(int i=0; i<(n); i++)
using ll = long long;
ll n;
set<ll> st;
int main(){
	scanf("%lld", &n);
	for(ll i=1; i*i <= n; i++) if(n % i == 0){
		ll k = n/i;
		rep(o,2){
			if(k == 1) continue;
			ll t = n;
			while(t % k == 0) t /= k;
			if(t % k == 1) st.insert(k);
			k = n / k;
		}
	}
	n--;
	for(ll i=1; i*i <= n; i++) if(n % i == 0){
		st.insert(i); st.insert(n/i);
	}
	printf("%lu\n", st.size() - 1);
}