Problem:
Score: 300 points

Problem Statement

As AtCoder Beginner Contest 100 is taking place, the office of AtCoder, Inc. is decorated with a sequence of length N, a = {a_1, a_2, a_3, ..., a_N}.

Snuke, an employee, would like to play with this sequence.

Specifically, he would like to repeat the following operation as many times as possible:

For every i satisfying 1 \leq i \leq N, perform one of the following: "divide a_i by 2" and "multiply a_i by 3".
Here, choosing "multiply a_i by 3" for every i is not allowed, and the value of a_i after the operation must be an integer.

At most how many operations can be performed?

Constraints

N is an integer between 1 and 10 \ 000 (inclusive).

a_i is an integer between 1 and 1 \ 000 \ 000 \ 000 (inclusive).

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 a_3 ... a_N

Output

Print the maximum number of operations that Snuke can perform.

Sample Input 1

3
5 2 4

Sample Output 1

3

The sequence is initially {5, 2, 4}. Three operations can be performed as follows:

First, multiply a_1 by 3, multiply a_2 by 3 and divide a_3 by 2. The sequence is now {15, 6, 2}.

Next, multiply a_1 by 3, divide a_2 by 2 and multiply a_3 by 3. The sequence is now {45, 3, 6}.

Finally, multiply a_1 by 3, multiply a_2 by 3 and divide a_3 by 2. The sequence is now {135, 9, 3}.

Sample Input 2

4
631 577 243 199

Sample Output 2

0

No operation can be performed since all the elements are odd. Thus, the answer is 0.

Sample Input 3

10
2184 2126 1721 1800 1024 2528 3360 1945 1280 1776

Sample Output 3

39

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();

		long[] arr = new long[n];
		for (int i = 0; i < n; i++) {
			arr[i] = sc.nextLong();
		}
		sc.close();

		solve(arr);

	}

	static void solve(long[] arr) {

		Boolean f = true;
		boolean s = true;
		int tc = 0;
		int count = 0;

		while (f) {

			if (!check(arr)) {
				break;
			}

			for (int i = 0; i < arr.length; i++) {
				long n = arr[i];

				if ((n % 2) != 0) {
					if (tc != arr.length - 1) {
						tc++;
					} else {
						f = false;
						break;
					}
				} else {
					if ((n / 2) < 1) {
						tc++;
					} else {
						if (s) {
							arr[i] = n / 2;
							s = false;
						}
					}
				}
			}
			tc = 0;
			s = true;
			count++;
		}

		System.out.println(count);
	}

	static boolean check(long[] arr) {

		long c = 0;
		for (int i = 0; i < arr.length; i++) {
			long n = arr[i];
			if ((n % 2) != 0) {
				c++;
			}
		}
		if (c == arr.length) {
			return false;
		}
		return true;
	}
}