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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(int argc, char *argv[]) {
    int n = 0;
    int i = 0;
    int j = 0;
    int c = 0;
    int count = 0;
    int d = 0;
    int max = 0;
    int maxv = 0;

    scanf("%d", &n);
    long a[n];

    while (n >= (i + 1))
    {
        scanf("%ld", &a[i]);
        i++;
    }
    i = 0;
    for (;;) {
        while (n >= (i + 1)) {//a[0]を最大値にする
            if ((a[i] > max) && (a[i] % 2 != 1)) {
               max = i;
               maxv = a[i];
            }
            i++;
        }
        i = max;
        if (i != 0) {
            while (i <= 1) {
                a[i] = a[i - 1];
                i--;
            }
        a[0] = maxv;
        }

        while (n >= (i + 1)) {
            if ((a[i] * 3) > 1000000000) {
                if(a[i] % 2 != 1) {
                    a[i] = a[i] / 2;
                    c = 1;
                } else {
                    d = 1;
                    break;
                }
            } else if (c == 0) {
                if((a[i] % 2) != 1) {
                    a[i] = a[i] / 2;
                    c = 1;
                } else {
                    a[i] = a[i] * 3;
                }
            } else {
                a[i] = a[i] * 3;
            }
            i++;
        }
//        printf("c:%d \n" , c);

        if (c != 1)
            break;
        if (d != 0)
            break;
        count++;
        c = 0;
//        i = 0;
//        while(n >= (i + 1)) {
//            printf("a[%d] = %d\n", i, a[i]);   /* 配列内容の表示 */
//            i++;
//        }
//        printf("\n");
        i = 0;
    }

    printf("%d\n", count);


    return 0;
}