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;
typedef long long ll;
typedef pair<int,int>P;
const double eps = 1e-8;
const int NINF = 0xc0c0c0c0;
const int INF  = 0x3f3f3f3f;
const ll  mod  = 1e9 + 7;
const ll  maxn = 1e6 + 5;

int main() {
	ll n;cin>>n;
	ll res=0;
	for(ll i=1;i*i<=n-1;i++){
		if((n-1)%i==0){
			if(i>1) res++;
			if((n-1)/i!=i) res++;
		}
	}
	auto check=[&](ll a){
		ll x=n;
		while(x%a==0) x/=a;
		if(x%a==1) res++;
	};
	for(ll i=1;i*i<=n;i++){
		if(n%i==0){
			if(i>1) check(i);
			if(i!=n/i) check(n/i);
		}
	}
	cout<<res<<'\n';
}