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 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<math.h>

int main(){
	long long n;
	long long cnt = 0;

	scanf("%lld", &n);

	long long sq = (long long)sqrt(n-1);
	long long k, tmp = n - 1;
	for(k=2; k<=sq; k++){
		if(tmp % k == 0){
			cnt += (k == sq) ? 1 : 2;
		}
	}
	cnt++; // n-1 % n-1 == 0

	sq = (long long)sqrt(n);
	for(k=2; k<=sq; k++){
		if(n % k != 0){
			continue;
		}

		tmp = n;
		while(tmp % k == 0){
			tmp/= k;
		}

		if(tmp % k == 1){
			cnt++;
		}
	}
	cnt++; // (n / n) % n == 1

	printf("%lld", cnt);

	return 0;
}