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<cstdio>
typedef long long ll;
int tot=0;
ll p[10005],sum[100005];
int main() {
	ll cnt=0,n,tmp;
	scanf("%lld",&n);
	tmp=n;
	for(register ll i=2;i*i<n;++i) {
		if((n-1)%i==0) {
			++cnt;
			if(i!=(n-1)/i) ++cnt;
		}
	}
	for(register ll i=2;i*i<=n;++i) {
		if(n%i==0) {
			tmp=n;
			while(tmp%i==0) tmp/=i;
			if(tmp%i==1) ++cnt;
			if(i*i==n) continue;
			tmp=n;
			while(tmp%(n/i)==0) tmp/=(n/i);
			if(tmp%i==1) ++cnt;
		}
	}
	if(n!=2) ++cnt;
	printf("%lld\n",cnt+1);
	return 0;
}