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>
int main()
{
	long long int n;
	scanf("%lld", &n);
	long long int ans = 1;
	long long int v, i;
	for (i = 2; i * i <= n; i++)
	{
		v = n;
		while (v % i == 0)
			v /= i;
		v %= i;
		if (v == 1)
			ans++;
	}
	long long int s = i - 1;
	for (i = 1; i * i <= n; i++)
	{
		if ((n - 1) % i == 0)
		{
			if ((n - 1) / i >= s)
				ans++;
			else
				break;
		}
	}
	printf("%lld\n", ans);
	return 0;
}