Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

6

Sample Output 1

3

There are three choices of K in which N becomes 1 in the end: 2, 5, and 6.

In each of these choices, N will change as follows:

When K=2: 6 \to 3 \to 1

When K=5: 6 \to 1

When K=6: 6 \to 1

Sample Input 2

3141

Sample Output 2

13

Sample Input 3

314159265358

Sample Output 3

9

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.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;}
	for(i=2;i*i<=n-1;i++)
		if((n-1)%i==0){
			cnt++;
			if(i!=(n-1)/i)cnt++;
		}
	cnt++;
	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++;
	printf("%lld\n",cnt);
	return 0;
}