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)

#include <iostream>
#include <string>
#include <vector>
#include <set>
#include <stack>
#include <queue>
#include <map>
#include <algorithm>
#include <iomanip>
#include <math.h>
#include <string.h>
#include <cstdio>
#include <tuple>
using namespace std; using ll = long long; using ld = long double;  using pll = pair<ll, ll>;
using vl = vector<ll>; using vll = vector<vl>; using vpll = vector<pll>;
const ld PI = 3.1415926535897932;
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#define rep1(n) for(ll i=0;i<n;++i)
#define rep2(i,n) for(ll i=0;i<n;++i)
#define rep3(i,a,b) for(ll i=a;i<b;++i)
#define rep4(i,a,b,c) for(ll i=a;i<b;i+=c)
#define rep(...) overload4(__VA_ARGS__,rep4,rep3,rep2,rep1)(__VA_ARGS__)
#define mp make_pair
#define mt make_tuple
#define INT(...) int __VA_ARGS__;in(__VA_ARGS__)
#define LL(...) ll __VA_ARGS__;in(__VA_ARGS__)
#define ULL(...) ull __VA_ARGS__;in(__VA_ARGS__)
#define STR(...) string __VA_ARGS__;in(__VA_ARGS__)
#define CHR(...) char __VA_ARGS__;in(__VA_ARGS__)
#define DBL(...) double __VA_ARGS__;in(__VA_ARGS__)
#define LD(...) ld __VA_ARGS__;in(__VA_ARGS__)
void in() {}
void debug_out() { cerr << endl; }
template<typename Head, typename... Tail>
void debug_out(Head h, Tail... t) {
	cerr << " " << h;
	if (sizeof...(t) > 0) cerr << " :";
	debug_out(t...);
}
template <typename T>
ostream& operator<<(ostream& os, vector<T> vec) {
	for (size_t i = 0; i < vec.size(); i++)os << vec[i] << (i + 1 == vec.size() ? "" : " ");
	return os;
}
ll ctoi(char c) {
	switch (c) {
	case '0': return 0; case '1': return 1; case '2': return 2;
	case '3': return 3; case '4': return 4; case '5': return 5;
	case '6': return 6; case '7': return 7; case '8': return 8;
	case '9': return 9;  default: return 0;
	}
}
void vout(vl V) {
	for (ll i = 0; i < (ll)V.size(); i++) cout << V[i] << " ";
	cout << endl;
}
bool pairCompare(const pll firstElof, pll secondElof)
{
	return firstElof.first > secondElof.first;
}
//**Snippetリスト**//
//rep, vin, all, iteLoop//
//bitSearch, bitList, nod, gcdlcm, isPrime, primeFactorize, Npow, divisor, modinv, dfs, bfs, eratos//
//dijkstra, WarshallFloyd, BellmanFord, UnionFind, COM, digitDP ,RMQ//
ll i, j, k, l; ll N, M, K, H, W, L, X, Y, Z, R, Q;
ll MOD = 1000000007; ll INF = 1LL << 62; ll ans = 0; ll z = 0, o = 1;
vl flag, color; vll path; vl A; vl dp; vl Ans;
int main(){
	cin >> N;
	vl A(N);
	for (i = 0; i < N; i++) cin >> A[i];
	for (i = 0; i < N; i++) {
		while (A[i] % 2 == 0) {
			A[i] /= 2;
			ans++;
		}
	}
	cout << ans;
}