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 <stdio.h>
#define LL long long
LL n,cnt;
int judge(LL k,LL m){
	while(m>=k){
		if(m%k==0)m/=k;
		else m%=k;
	}
	if(m==1)return 1;
	else return 0;
}
int main(){
	LL i;
	scanf("%lld",&n);
	if(n==2){printf("1\n");return 0;}//注意n=2需特判
	for(i=2;i*i<=n-1;i++)//找n-1的约数
		if((n-1)%i==0){
			cnt++;
			if(i!=(n-1)/i)cnt++;
		}
	cnt++;//加上n-1本身
	for(i=2;i*i<=n;i++)
		if(n%i==0){
			if(judge(i,n))cnt++;
			if(i!=n/i&&judge(n/i,n))cnt++;
		}
	cnt++;//加上n本身
	printf("%lld\n",cnt);
}