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 (Clang 3.8.0) (judge verdict: Accepted)

#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <inttypes.h>
#include <stdint.h>
#include <string.h>
#include <wchar.h>

#define MIN(a, b) (a < b ? a : b)
#define MAX(a, b) (a > b ? a : b)

uint32_t a[10000];

int main(void) {
	uint32_t N;

	scanf("%"PRIu32"\n", &N);
	for (uint32_t idx = 0; idx < N - 1; idx++) {
		scanf("%"PRIu32" ", &(a[idx]));
	}
	scanf("%"PRIu32, &(a[N - 1]));

	uint32_t min2Div = UINT32_MAX;
	uint32_t max2Div = 0;
	uint32_t min3Mul = UINT32_MAX;
	uint32_t max3Mul = 0;
	uint32_t minOpCount = UINT32_MAX;
	uint32_t div2Sum = 0;
	uint32_t mul3Sum = 0;

	for (uint32_t idx = 0; idx < N; idx++) {
		uint32_t div2Count = 0;

		while (a[idx] % 2 == 0 && a[idx] > 1) {
			div2Count++;
			a[idx] /= 2;
		}
		div2Sum += div2Count;

		uint32_t mul3Count = 0;
		while (a[idx] <= UINT32_C(1000000000) / UINT32_C(3)) {
			mul3Count++;
			a[idx] *= 3;
		}
		mul3Sum += mul3Count;

		if (div2Count + mul3Count < minOpCount) {
			minOpCount = div2Count + mul3Count;
		}

		if (div2Count < min2Div) {
			min2Div = div2Count;
		}
		if (max2Div < div2Count) {
			max2Div = div2Count;
		}
		if (mul3Count < min3Mul) {
			min3Mul = mul3Count;
		}
		if (max3Mul < mul3Count) {
			max3Mul = mul3Count;
		}
	}

	printf("%"PRIu32,div2Sum);

	return 0;
}