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 <vector>
#include <string>
#include <utility>
#include <queue>
#include <algorithm>
using namespace std;

#define ll long long

ll MAX(ll a, ll b) {

	if (a > b)return a;
	else return b;

}
ll MIN(ll a, ll b) {

	if (a < b)return a;
	else return b;

}

ll powsur(ll a, ll b, ll p) {

	if (b == 0) {

		return 1;

	}
	else if (b % 2 == 0) {

		ll d = powsur(a, b / 2, p);
		return ((d%p)*(d%p)) % p;

	}
	else {

		return ((a%p)*powsur(a, b - 1, p) % p) % p;

	}

}
//pは素数,aのb乗をpで割った余り

ll CMOD(ll n, ll r, ll p) {

	ll a = 1, b = 1, c = 1, re;
	for (int i = 1; i <= n; i++) {

		a *= i;
		a = a%p;

	}
	for (int i = 1; i <= r; i++) {

		b *= i;
		b = b%p;

	}
	for (int i = 1; i <= n - r; i++) {

		c *= i;
		c = c%p;

	}
	b = powsur(b, p - 2, p);
	c = powsur(c, p - 2, p);
	re = ((a%p)*(b%p)) % p;
	re = ((re%p)*c%p) % p;

	return re;

}
//pは素数,(nCr)%p

ll gcd(ll a, ll b) {

	if (a < b) {

		ll tmp = a;
		a = b;
		b = tmp;

	}
	if (a%b == 0)return b;
	return gcd(b, a%b);

}
//最大公約数

ll lcm(ll a, ll b) {

	return a / gcd(a, b)*b;

}
//最小公倍数

ll func(ll num) {

	ll cnt = 0;
	while (num % 2 == 0) {
		cnt++;
		num /= 2;
	}
	return cnt;

}

int main() {

	ll N, cnt = 0;
	cin >> N;
	for (int i = 0; i < N; i++) {

		int a;
		cin >> a;
		cnt += func(a);

	}
	cout << cnt << endl;

	return 0;
}