Problem: Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.

Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.

There are two conditions though: 

  * For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x. 
  * The modified rating changes must be perfectly balanced — their sum must be equal to 0. 



Can you help with that?

Input

The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.

Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.

The sum of all a_i is equal to 0.

Output

Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.

For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.

If there are multiple solutions, print any. We can show that a solution exists for any valid input.

Examples

Input


3
10
-5
-5


Output


5
-2
-3


Input


7
-7
-29
0
3
24
-29
38


Output


-3
-15
0
2
12
-15
19

Note

In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.

In the second example there are 6 possible solutions, one of them is shown in the example output.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
bool isLucky(int n) {
  bool arr[10];
  for (int i = 0; i < 10; i++) arr[i] = false;
  while (n > 0) {
    int digit = n % 10;
    if (arr[digit]) return false;
    arr[digit] = true;
    n = n / 10;
  }
  return true;
}
bool isPrime(int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
int main() {
  int n;
  cin >> n;
  int m = n;
  int arr[n];
  int cnt1 = 0;
  int cnt2 = 0;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  int sum = 0;
  int ans = 2;
  vector<pair<int, int>> v;
  for (int i = 0; i < n; i++) {
    if (arr[i] % 2 != 0) {
      v.push_back(make_pair(ceil(double(arr[i]) / 2), i));
    }
    sum += ceil(double(arr[i]) / 2);
  }
  int i = 0;
  while (sum != 0) {
    v[i].first = floor(double(arr[v[i].second]) / 2);
    i++;
    if (sum > 0) {
      sum--;
    } else
      sum++;
  }
  int j = 0;
  for (int i = 0; i < n; i++) {
    if (arr[i] % 2 == 0) {
      cout << arr[i] / 2 << endl;
    } else {
      cout << v[j].first << endl;
      j++;
    }
  }
}