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.

Examples

Input

6


Output

3


Input

3141


Output

13


Input

314159265358


Output

9
from math import sqrt, ceil


def divisors(n):
    out = []
    nn = ceil(sqrt(n))
    for i in range(1, nn):
        if n % i == 0:
            out.append(i)
            out.append(n//i)

    if nn ** 2 == n:
        out.append(nn)

    out.sort()
    return out


n = int(input())
a = len(divisors(n-1)[1:])

d = divisors(n)
for dd in d[1:]:
    nn = n
    while nn % dd == 0:
        nn = nn // dd
    if nn % dd == 1:
        a += 1

print(a)
