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 Kotlin (1.0.0) (judge verdict: Accepted)

fun main(args: Array<String>) {
    val n = readLine()?.toInt() ?: return
    val aList = readLine()?.split(" ")?.map { it.toLong() } ?: return

    // ここで、操作制限があるのは 2で割る ことであることに着目すると,
    // 配列中の偶数要素の個数が問題になる.
    // n = x * 2^i という n を整数の範囲で 3倍、1/2、3倍、1/2...
    // と操作できる回数は i　回である.
    // 上記 n については、最初に何回3倍してもこの性質は崩れない
    // 戦略としては任意の2の倍数 n についてまず1/2操作、それ以外の要素にたいして3倍操作を行い
    // nが2の倍数ではなくなったら、別の2の倍数を選んで同じことを繰り返すのが最適
    // よって、n = x * 2^i の i を合計したものが、操作回数の最大値となる

    val result = aList.filter { it % 2L == 0L }
        .map {
            var i = 0
            var half = it
            while (half % 2L == 0L) {
                half /= 2
                i++
            }
            i
        }
        .sum()

    println(result)
}