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 rep(i,n) for(int i=0; i<(int)(n); i++)

using namespace std;
typedef long long LL;

int main(){
	LL N;
	cin >> N;
	LL ans=1, res=0;
	LL x[1000010]={0};
	LL num = N-1;
	for(LL i=2; i*i<=N-1; i++){
		while(num%i==0){
			x[i]++;
			num/=i;
		}
	}
	for(LL i=2; i*i<=N-1; i++) ans*=x[i]+1;
    if(num!=1) ans*=2;
	for(LL K=2; K*K<=N; K++){
		LL Nc=N;
		while(Nc%K==0) Nc/=K;
		if(Nc%K==1) res++;
	}
	cout << res+ans/2+1 << endl;

	return 0;
}
