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;
template <typename T>
static ostream& operator<<(ostream& stream, const vector<T>& t);
template <typename T>
static istream& read(T, T, istream& = cin);
template <typename T>
static istream& operator>>(istream& stream, vector<T>& t) {
  return read((t).begin(), (t).end(), stream);
}
template <typename T>
static istream& read(T b, T e, istream& stream) {
  for (T it = b; it != e; ++it) stream >> *it;
  return stream;
}
template <typename T>
static T sum(vector<T>& arr) {
  T ans = 0;
  for (auto x : arr) ans += x;
  return ans;
}
struct _print {
  string sep, end;
  bool space;
  ostream& stream;
  _print(string _sep = " ", string _end = "\n", ostream& _stream = cout)
      : sep(_sep), end(_end), space(false), stream(_stream) {}
  ~_print() { stream << end; }
  template <typename T>
  _print& operator,(const T& t) {
    if (space) stream << sep;
    space = true;
    stream << t;
    return *this;
  }
};
struct _print2 {
  string s1, s2, end;
  bool space;
  ostream& stream;
  _print2(string _s1 = " ", string _s2 = " ", string _end = "\n",
          ostream& _stream = cout)
      : s1(_s1), s2(_s2), end(_end), space(false), stream(_stream) {}
  ~_print2() { stream << end; }
  template <typename T>
  _print2& operator,(const T& t) {
    for (auto _ : t) {
      if (space) stream << s1;
      space = true;
      _print p(s2, "", stream);
      p, _;
    }
    return *this;
  }
};
void __() {
  int n;
  cin >> n;
  vector<long long> a(n);
  cin >> a;
  vector<long long> b(n);
  for (int i = 0; i < n; ++i) b[i] = (a[i] - bool(a[i] % 2)) / 2;
  long long x = -sum(b);
  for (int i = 0; i < n; ++i)
    if (a[i] % 2)
      if (x > 0) {
        --x;
        b[i]++;
      }
  _print2("\n", " ", "\n"), b;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << fixed << setprecision(15);
  __();
}