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: Accepted)

#include<stdio.h>
int main()
{
	long long int n;
	scanf("%lld", &n);
	long long int ans = 1;
	long long int v, i;
	if (n < 100)
	{
		for (i = 2; i < n; i++)
		{
			v = n;
			while (v % i == 0)
				v /= i;
			v %= i;
			if (v == 1)
				ans++;
		}
		printf("%lld\n", ans);
	}
	else
	{
		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;
}