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 <stdbool.h>
#include <string.h>
#include <time.h>
typedef long long int ll;
#define mod 1000000007
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define m0(x) memset(x,0,sizeof(x))


//組み合わせ（nとrが少ないとき）
int combination_low(ll n,ll r){
	const ll N = 500;
	ll C[N][N];
	memset(C, 0, sizeof(C));
	C[0][0]=1;
	for(ll n=1; n<N; n++){
		C[n][0]=C[n-1][0];
		for(ll r=1; r<N; r++){
			C[n][r] = C[n-1][r-1] + C[n-1][r];
		}
	}
	return C[n][r];
}

//二分累乗法
int powa(ll a, ll n){
	ll res = 1;
	while(n>0) {
		if(n&1) res = res * a % mod;
		a = a * a % mod;
		n >>= 1; 
	}
	return res;
}

//逆元
int inverse(ll a){
	ll n = mod-2;
	return powa(a, n);
}

/*組み合わせ（いちばんうまくいったやつ）*/
int combination(ll n,ll r){
	ll x=1,y=1;
	for(int i=1;i<=r;i++){
		x = x * (n-i+1) % mod;
		y = y * i % mod;
	}
	return x * inverse(y) % mod;
}

int main(void){
	int n;
	scanf("%d",&n);
	ll a[n];
	int count=0;
	rep(i,n){
		scanf("%d",&a[i]);
		while((a[i]%2)==0){
			a[i]/=2;
			count++;
		}
	}
	printf("%d\n",count);
	return 0;

}