Ivan has number b. He is sorting through the numbers a from 1 to 10^{18}, and for every a writes ([a,    b])/(a) on blackboard. Here [a,    b] stands for least common multiple of a and b. Ivan is very lazy, that's why this task bored him soon. But he is interested in how many different numbers he would write on the board if he would finish the task. Help him to find the quantity of different numbers he would write on the board.

Input

The only line contains one integer — b (1 ≤ b ≤ 10^{10}).

Output

Print one number — answer for the problem.

Examples

Input

1


Output

1

Input

2


Output

2

Note

In the first example [a,    1] = a, therefore ([a,    b])/(a) is always equal to 1.

In the second example [a,    2] can be equal to a or 2 ⋅ a depending on parity of a. ([a,    b])/(a) can be equal to 1 and 2.
from sys import stdin, stdout
import cProfile, math
from collections import Counter
from bisect import bisect_left
import itertools

printHeap = str()
test = False
memory_constrained = False
N = 10**9+7
import sys
sys.setrecursionlimit(10000000)


def display(string_to_print):
    stdout.write(str(string_to_print) + "\n")


def primeFactors(n): #n**0.5 complex 
    factors = dict()
    for i in range(2,math.ceil(math.sqrt(n))+1):  
        while n % i== 0: 
            if i in factors:
                factors[i]+=1
            else: factors[i]=1
            n = n // i 
    if n>2:
        factors[n]=1
    return (factors)
     
def isprime(n):
    """Returns True if n is prime."""
    if n < 4:
        return True
    if n % 2 == 0:
        return False
    if n % 3 == 0:
        return False
    i = 5
    w = 2
    while i * i <= n:
        if n % i == 0:
            return False
        i += w
        w = 6 - w
    return True

def test_print(*args):
    if test:
        print(args)


def display_list(list1, sep=" "):
    stdout.write(sep.join(map(str, list1)) + "\n")


def get_int():
    return int(stdin.readline().strip())


def get_tuple():
    return map(int, stdin.readline().split())


def get_list():
    return list(map(int, stdin.readline().split()))


memory = dict()


def clear_cache():
    global memory
    memory = dict()


def cached_fn(fn, *args):
    global memory
    if args in memory:
        return memory[args]
    else:
        result = fn(*args)
        memory[args] = result
        return result


# ----------------------------------------------------------------------------------- MAIN PROGRAM
TestCases = False

def main():
    b = get_int()
    if b==1:
        print(1)
    else:
        if isprime(b): 
            print(2)
            return
        res = 1
        factors = primeFactors(b)
        for i in factors.keys():
            res = (factors[i]+1)*res
        print(res)

    

# --------------------------------------------------------------------------------------------- END



if TestCases: 
    for _ in range(get_int()): 
        cProfile.run('main()') if test else main()
else: cProfile.run('main()') if test else main()