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;
namespace QiFeng233{
	ll n,sn,ans;
	bool chk(ll x){
		ll tmp=n;
		while(tmp>1){
			if(tmp%x==0)tmp/=x;
			else{
				if((tmp-1)%x==0)return 1;
				else return 0;
			}
		}
		if(tmp==1)return true;
		else return false;
	}
	void solve(){
		scanf("%lld",&n);
		sn=(ll)sqrt((double)n);
		for(ll i=2;i<=sn;++i){
			if(chk(i))++ans;
		}
		for(ll i=1;i<=sn;++i)if((n-1)%i==0)++ans;
		if(n-1==sn*sn)--ans;
		printf("%lld",ans+1);
	}
}
signed main(){
	QiFeng233::solve();
	return 0;
}