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: Accepted)

    #include <stdio.h>
    #include <string.h>
     

    int main();
    int operation();
    int judge();

    int main(){
        int count = 0;
        int N;
        int r = 0;
        int j = 1;
  //      printf("%s","Nを入力してください");
         scanf("%d",&N);
   
    int a[N];
    int i = 0;
  //   printf("%s","数列を入力してください\n");
     while(i< N){
          scanf("%d",&a[i]);
          i++;
        }

 
    
    while(j>0){
        i = 0;
        j = 0;
        while(i<N){
            r = r + judge(a[i]);
            a[i] = operation(a[i]);
            j = j + judge(a[i]);
            i++;
        }
    } 
    

    printf("%d%s",r,"\n");
    
}
    
int operation(int a){
      if(a%2 == 0){
          a = a/2;
      }else{
          a = a*3;
      }

     return a;
    
 }

int judge(int a){
    int r = 0;
    if(a%2 == 0){
        r = 1;
    }
    return r;
}