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 Python (3.4.3) (judge verdict: Accepted)

#約数全列挙
def make_divisors(n):
	divisors = []
	for i in range(1, int(n**0.5)+1):
		if n % i == 0:
			divisors.append(i)
			if i != n // i:
				divisors.append(n//i)
	return divisors


n=int(input())
#n%k!=0
cnt=len(make_divisors(n-1))-1

#n%k==0
lst=make_divisors(n)
for k in lst:
	if k!=1:
		t=n
		while t%k==0:
			t//=k
		if (t-1)%k==0:
			cnt+=1
print(cnt)