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 <iostream>
#include <vector>
#include <algorithm>

using namespace std;

bool check(long long n, long long d)
{
    while(n%d==0) n/=d;
    n=n%d;
    return n==1;
}

int main()
{
    long long n; cin>>n;
    long long ans=0;
    for(long long d=1;d*d<=n;++d)
    {
        if(n%d) continue;
        if(d>1) ans+=check(n,d);
        if(d*d!=n) ans+=check(n, n/d);
    }
    for(long long d=1;d*d<n;++d)
    {
        if((n-1)%d) continue;
        if(d>1) ans++;
        if(d*d<n-1) ans++;
    }
    cout << ans << "\n";
}