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>
#define int long long
using namespace std;
int n,ans=0;
map<int,bool> visited;
map<int,bool>::iterator it;
inline void get(int x)
{
	if (x==1)  return;
	int m=n;
	while (m%x==0) m/=x;
	if (m%x==1 && !visited[x]) visited[x]=1;
}
signed main()
{
	cin>>n;
	for (int i=1;i*i<=n;i++) if (n%i==0)  get(i),get(n/i);
	for (int i=1;i*i<=n-1;i++) if ((n-1)%i==0)  get(i),get((n-1)/i);
	for (it=visited.begin();it!=visited.end();it++)  ans++;
	cout<<ans<<endl;
	return 0;
}