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 pf printf
#define sc(x) scanf("%d",&x);
#define scs(x) scanf("%s",x);
#define scl(x) scanf("%lld",&x);
#define rep(i,s,e) for(int i=s; i<=e; ++i)
#define dep(i,s,e) for(int i=s; i>=e; --i)
typedef long long ll;
using namespace std;
const ll mod=998244353;
int main()
{
	ll n,cou=2;
	scl(n);
	if(n==2){
		pf("1\n");return 0;
	}
	for(ll i=2;i*i<=n;i++){
		ll u=n;
		if(u%i==0){
			while(u%i==0)u/=i;
			if(u%i==1)cou++;
		}else {
			if(u%i==1)cou+=2;
			if(i*i==n-1)cou--;
		}
	}
	pf("%lld\n",cou);
}
