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>
const long long INF = 1e18, mod = 998244353, MAXN = 1e5 + 2, SQ = 322;
const long double EPS = 1e-6, PI = 3.141592653589793238462643;
using namespace std;
namespace in_out {
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> a) {
  out << a.first << " " << a.second;
  return out;
}
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T> a) {
  for (auto i : a) out << i << " ";
  out << '\n';
  return out;
}
template <typename T>
istream& operator>>(istream& in, vector<T>& a) {
  for (long long i = 0; i < a.size(); ++i) in >> a[i];
  return in;
}
}  // namespace in_out
using namespace in_out;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> a(n), b;
  cin >> a;
  b = a;
  vector<bool> u(n);
  for (int i = 0; i < n; i++)
    if (a[i] % 2) u[i] = 1;
  for (auto& i : a) i /= 2;
  int sum = 0;
  for (auto& i : a) sum += i;
  if (sum > 0) {
    for (int i = 0; i < n; i++) {
      if (b[i] < 0 && u[i]) a[i]--, sum--;
      if (sum == 0) break;
    }
  } else if (sum < 0) {
    for (int i = 0; i < n; i++) {
      if (b[i] > 0 && u[i]) a[i]++, sum++;
      if (sum == 0) break;
    }
  }
  cout << a;
}