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: Wrong Answer)

from sys import stdin
def make_divisors(n):
    mod0 = set()
    mod1 = set()
    for i in range(2, int(n**0.5)+1):
        if n%i==0:
            mod0.add(i)
            mod0.add(n/i)
        if (n-1)%i==0:
            mod1.add(i)
            mod1.add((n-1)/i)
    print(mod0)
    print(mod1)
    return mod0,mod1

N=list(map(int,(stdin.readline().strip().split())))
num=N[0]
K=0
mod0,mod1=(make_divisors(num))
# mod0.remove(1)
# mod1.remove(1)
for i in mod0:
    num = N[0]
    while(num % i == 0):
        num/=i

    if num%i==1:
        K+=1
K+=len(mod1)+2
print(K)