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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#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;
	//printf("1\n");
	for(register ll i=1;i*i<n;++i) {
		if((n-1)%i==0) {
			++cnt;
			if(i!=(n-1)/i&&i!=1) ++cnt;
		}
	}
	//printf("%lld\n",cnt);
	for(register ll i=2;i*i<=tmp;++i) {
		if(tmp%i==0) {
			p[++tot]=i;sum[tot]=1;
			while(tmp%i==0) tmp/=i,sum[tot]*=i;
		}
	}
	if(tmp>1) {p[++tot]=tmp;sum[tot]=tmp;}
	for(register ll i=1;i<=tot;++i) {
		if(n%sum[i]==0) {
			//printf("p:%d\n",p[i]);
			tmp=n/sum[i];
			cnt+=(tmp%p[i]==1);
		}
	}
	printf("%lld\n",cnt+1);
	return 0;
}