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;
bool can(ll x, ll p) {
  while(x % p == 0) {
    x /= p;
  }
  return x % p == 1;
}
set<ll>s;
int main()
{
    ll n;
    scanf("%lld",&n);
    ll m=n-1;
    for(ll i=1;i*i<=m;i++)
    {
        if(m%i==0)
        {
            s.insert(m/i);
            s.insert(i);
        }
    }
    for(ll i=1;i*i<=n;i++)
    {
        if(i != 1) if(can(n, i)) s.insert(i);
      if((n / i) != 1) if(can(n, n / i)) s.insert(n / i);
    }
    s.erase(1);
    printf("%d\n",s.size());
}
