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: Runtime Error)

#include <stdio.h>

typedef struct{
	int max;
	int order;
}MAX;

void max_order(int a[], int size, MAX *ret){
	int max, order;
	max=a[size];
	while(size){
		size--;
		if(max<a[size]){
			max=a[size];
			order=size;
		}
	}
	ret->max=max;
	ret->order=order;
}

int main(){
	int N, i, j, chain=0, ord;
	MAX data;
	scanf("%d", &N);
	int a[N], b[N], c;
	for(i=0;i<N;i++){
		scanf("%d", &a[i]);
	}
	for(i=0;i<N;i++){
		c=a[i];
		b[i]=0;
		while(c%2==0){
			b[i]++;
			c/=2;
		}
	}
	for(i=0;;i++){
		max_order(b, N, &data);
		if(data.max==0){
			break;
		}
		a[data.order]/=2;
		b[data.order]--;
		for(j=0;j<data.order;j++){
			a[j]*=3;
		}
		for(j=data.order+1;j<N;j++){
			a[j]*=3;
		}
	}
	printf("%d", i);
}