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>

void arr_input(long long int arr[], int n);
void arr_output(long long int arr[], int n);
int count_3_or_2(long long int arr[], int n);
int odd_check(long long int arr[], int odd[], int n);

int main(void) {

  int n;
  long long int arr[10010];
  scanf("%d", &n);

  arr_input(arr, n);
  printf("%d\n", count_3_or_2(arr,n));
  
  return 0;
}

int count_3_or_2(long long int arr[], int n){

  int count = 0;
  int odd[10010] = {0};
  int k;
  int odd_f;
  int first = 0;

  while(1) {
    odd_f = odd_check(arr,odd,n);
    if (odd_f == 1) { break;}

    for (k = 0; k < n; k++) {
      if (odd[k] == 0) {
	if (first == 0) {
          arr[k] /= 2;
	  first = 1;
	} else {
	  if (arr[k] *3 > 1000000000) {
	    arr[k] /= 2;
	  } else {
            arr[k] *= 3;
	  }
	}	
      } else {
        arr[k] *= 3;
      }
    }
    arr_output(arr,n);
    first = 0;
    count++;
  }


 
  return count;
}

int odd_check(long long int arr[], int odd[], int n) {

  int k;
  int odd_c = 0;
  
  for (k = 0; k < n; k++) {
    if (arr[k] % 2 == 1) {
      odd[k] = 1;
      odd_c++;
    }
  }

  if (odd_c == n) {
    return 1;
  } 
  return 0;

}

void arr_input(long long int arr[], int n) {

  int k;
  long long int tmp;

  for (k = 0; k < n; k++) {
    scanf("%lld", &tmp);
    arr[k] = tmp;
  }

}
void arr_output(long long int arr[], int n) {

  int k;

  for (k = 0; k < n; k++) {
    printf("%lld ", arr[k]);
  }
  puts("");

}