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)

n = int(input())
if n == 2:
    print(1)
    exit()
ans = []
divisors = []
for i in range(2,int(n**0.5)+1):
    #n-1の約数は全部
    if (n-1) % i == 0:
        ans.append(i)
        if i != (n-1) // i:
            ans.append((n-1)//i)
    #nの約数はやってみる
    if n % i == 0:
        divisors.append(i)
        if i != n // i:
            divisors.append(n//i)
for item in divisors:
    num = n
    while num%item == 0:
        num = num//item
    if num%item == 1:
        ans.append(item)
#print(divisors)
#print(ans)
print(len(ans)+2)#k=n,n-1がはいっていないので+2

'''
import math
n = int(input())
ans = []
divisors = []

for i in range(2,math.ceil(n**0.5)+1):
    num = n
    while num >= i:
        if num%i == 0:
            num //= i
        else:
            num -= i
    print(num)
    if num == 1:
        ans.append(i)
    if n % i == 0:
        divisors.append(i+1)
        if i != n // i:
            divisors.append(n//i+1)

print(ans)
print(divisors)
print(len(ans)+len(divisors))

#[2, 3, 4, 5, 10, 20, 157, 314, 628, 785, 1570, 3140, 3141]
'''