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>
#define fs first
#define se second
#define pb push_back
using namespace std;

typedef long long ll;
typedef pair<int,int> pii;
typedef double db;

const int maxn=5e5+5;
const ll inf=0x3f3f3f3f;
const ll mod=1e9+7;

char s[maxn];

ll n;
int check(ll num){
	ll tmp=n;
	while(tmp%num==0) tmp/=num;
	if(tmp%num==1){
		return 1;
	}
	else return 0;
}

int main(){	
	scanf("%lld",&n);
	ll ans=2;	
	if(n==2){
		printf("1");return 0;
	}
	for(ll i=2;i*i<=n;i++){
		ans+=check(i);
		if(i*i!=n&&n/i!=i) ans+=check(n/i);
	}
	printf("%lld",ans);
	return 0;
}