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 collections import Counter
import io
import os
import sys
from atexit import register
import random
import math
import itertools

##################################### Flags          #####################################
# DEBUG = True
DEBUG = False
STRESSTEST = False
# STRESSTEST = True

##################################### IO             #####################################

if not DEBUG:
    input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
sys.stdout = io.BytesIO()
register(lambda: os.write(1, sys.stdout.getvalue()))

tokens = []
tokens_next = 0


def nextStr():
    global tokens, tokens_next
    while tokens_next >= len(tokens):
        tokens = input().split()
        tokens_next = 0
    tokens_next += 1

    if type(tokens[tokens_next - 1]) == str:
        return tokens[tokens_next - 1]
    return tokens[tokens_next - 1].decode()


def nextInt():
    return int(nextStr())


def nextIntArr(n):
    return [nextInt() for i in range(n)]


def print(*argv, end='\n'):
    for arg in argv:
        sys.stdout.write((str(arg) + ' ').encode())
    sys.stdout.write(end.encode())


##################################### Helper Methods #####################################
def genTestCase():
    return random.randint(1, 100)


def lcm(a, b):
    return a * b // math.gcd(a, b)


def bruteforce(b):
    res = set()
    for a in range(1, 10**5):
        res.add(lcm(a, b) // a)
    return len(res)


def doStressTest():
    while True:
        curTest = genTestCase()
        mySoln = solve(curTest)
        bruteforceSoln = bruteforce(curTest)
        if mySoln != bruteforceSoln:
            print('### Found case ###')
            print(curTest)
            print(f'{mySoln} should have been: {bruteforceSoln}')
            return


def genPrimes(n):
    isPrime = [0, 0] + [1] * (n + 10)
    for i in range(len(isPrime)):
        if isPrime[i]:
            yield i
            for j in range(i * i, len(isPrime), i):
                isPrime[j] = False


primes = [i for i in genPrimes(10**5)]


def factorize(n):
    c = Counter()
    primesIter = iter(primes)
    while n > 1:
        p = next(primesIter, -1)
        if p == -1:
            c[n] += 1
            break
        while n % p == 0:
            c[p] += 1
            n //= p
    return c


def solve(b):
    f = factorize(b)
    res = 1
    for i in f:
        res *= f[i] + 1
    return res


##################################### Driver         #####################################

if __name__ == "__main__":
    if not DEBUG and STRESSTEST:
        raise Exception('Wrong flags!')

    if STRESSTEST:
        doStressTest()
    else:
        ### Read input here
        res = solve(nextInt())
        print(res)

    sys.stdout.flush()