Write a function:
def solution(N):

that, given a positive integer N, returns the number of its even factors.

For example, given N = 16, the function should return 3, because among all factors of 16 (1, 2, 4, 8, 16), only three factors (2, 4, 8) are even numbers.
Here's an inefficient solution:

def solution(N):
    count = 0
    # Check all numbers from 1 to N
    for i in range(1, N + 1):
        # Check if i is a factor of N
        if N % i == 0:
            # Check if the factor is even
            if i % 2 == 0:
                count += 1
    return count

The above solution works correctly but is highly inefficient for large numbers as it:

- Checks every number from 1 to N
- Makes unnecessary iterations
- Has a time complexity of O(N)

Write an efficient algorithm for the following assumptions:

- N is an integer within the range [1..2,147,483,647]