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;

ll yakusuu(ll n){
    if(n==1) return 0;
    ll ans=1;
    for(ll i=2;i*i<=n;i++){
        if(n%i==0){
            if(i*i!=n) ans+=2;
            else ans++;
        }
    }
    return ans;
}
int main(void){
    ll n;
    cin>>n;
    ll ans=yakusuu(n-1)+1;
    for(ll k=2;k*k<=n;k++){
        if(n%k==0){
            ll z=n;
            while(z%k==0){
                z/=k;
            }
            if(z%k==1){
                ans++;
            }
        }
    }
    cout<<ans;
}