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++14 (GCC 5.4.1) (judge verdict: Accepted)

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <iomanip>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <cmath>
#include <set>
#include <map>
#include <string>
#include <numeric>
#include <cstdio>
#include <cstdlib>
#include <ctime>
#include <iterator>
#include <fstream>
#include <cstddef>
#include <cstring>
//#include <chrono> //正確な時間計測のため
using namespace std;

typedef pair<int, int> P;
typedef long long ll;
#define PB push_back
#define rep(i, a) for(int i = 0; i < a; i++)
#define all(v) v.begin(),v.end()
#define FF first
#define SS second
#define mp make_pair
#define INT int_64
#define MOD 998244353
const ll INF = (1<<30);
const ll NEGA=-INF*10000; 

//階乗
ll Kaijou(ll a) {
if (a < 2) return 1;
return a * Kaijou(a - 1)%MOD;
}

//逆元の計算用(pは素数のときb=p-2で逆元)
ll calc(ll a, ll b, ll p) {
	if (b == 0) return 1;
	if (b % 2 == 0) {
		ll d = calc(a, b / 2, p);
		return (d*d) % p;
	}
	else {
		return (a*calc(a, b - 1, p)) % p;
	}
}


//ユークリッドの互除法
ll gcd(ll x, ll y) {
	if (x < y) swap(x, y);
	if (y == 0) return x;
	return gcd(y, x%y);
}

ll CCC(ll all,ll a,ll b,ll c){
  all=Kaijou(all)%MOD;
  a=Kaijou(a);
  b=Kaijou(b);
  c=Kaijou(c);
  ll sita=((a*b)%MOD*c)%MOD;
  return all*calc(sita,MOD-2,MOD)%MOD;
}

 ll ans=INF;

 ll cc[501][501];
 ll D[31][31];
 
 ll pdmin[3];

//再帰なし累乗
ll iPOW(ll x, ll p){
  ll result;
	for(result=1;p!=0;p>>=1)
	{
    if(p&1) result*=x;
	  x*=x;
  }
  return result;
}


int main(){  
 int n;
 cin>>n;
 int a[10000]={0};
 int ans=0;
 rep(i,n) {
   cin>>a[i];
   while(a[i]%2==0){
     ans++;
     a[i]/=2;
   }
 }
 cout<<ans<<endl;

}