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 PyPy3 (2.4.0) (judge verdict: Accepted)

n=int(input())
ans=0
def count_cases(n):
    n-=1
    cur=1
    cnt=0
    while cur*cur<=n:
        if n%cur==0:
            if n//cur!=cur:
                if (n+1)%(n//cur)==1:
                    cnt+=1
#                    print(n//cur)
                if (n+1)%cur==1:
                    cnt+=1
#                    print(cur)
            else:
                if (n+1)%cur==1:
                    cnt+=1
#                    print(cur)
        cur+=1
    return cnt
#count non divisible cases 
ans+=count_cases(n)
#print(ans,end=' ans\n')
#exit()
#count divisible type cases
cur=2
while cur*cur<=n:
    #check if cur,cur**2,cur**3,cur**4 are a valid one
    if n%cur==0:
        num=n
        div=cur
        while num%div==0:
            num//=div
        if num%div==1:
 #           print(div)
            ans+=1
        div=n//cur
        if cur!=(n//cur):
            num=n
            while num%div==0:
                num//=div
 #              print(div)
            if num%div==1:
                ans+=1
    cur+=1
print(1+ans)#extra 1 due to k=n case