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>
typedef long long int ll;
using namespace std;
#define endl '\n'
int main(){
    ll n;
    cin>>n;
    ll c=1;
    for(ll i=2;i*i<=n;i++){
        ll j=i;
        ll x=n;
        while(x%j==0)x/=j;
        if(x==1 || x%j==1)c++;
    }
    ll y=sqrt(n);
    for(ll i=1;i*i<=(n-1);i++){
        ll x=n-1;
        if(x%i==0){
            ll p=x/i;
            ll q=i;
            if(p==q && p>y)c++;
            else{
                if(p>y)c++;
                if(q>y)c++;
            }
        }
    }
    cout<<c<<endl;
    return 0;
}
