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;
#define rep(i,n) for(int i=0;i<n;i++)
#define all(v) v.begin(), v.end()
typedef long long ll;
typedef pair<ll,ll> P;
typedef vector<ll> vec;
typedef vector<vec> mat;
int main(){
    ll n,ans=0;
    cin>>n;
    for(ll i=2;i<=sqrt(n);i++){
        if(n%i==0){
            ll p=n;
            while(p%i==0) p/=i;
            if(p%i==1) ans++;
        }
        else if(n%i==1) ans++;
    }
    for(ll i=1;i<sqrt(n-1);i++){
        if((n-1)%i==0) ans++;
    }
    cout<<ans+1;
}