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;
inline void pisz(int n) { printf("%d\n", n); }
inline int rectCount(int n, int m) { return (m * n * (n + 1) * (m + 1)) / 4; }
vector<int> rem(vector<int> vect, int num) {
  vector<int> myVector = vect;
  std::vector<int>::iterator position =
      std::find(myVector.begin(), myVector.end(), num);
  if (position != myVector.end()) {
    myVector.erase(position);
  }
  return myVector;
}
const int fx[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
const int fxx[8][2] = {{0, 1}, {0, -1}, {1, 0},  {-1, 0},
                       {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
template <typename T, typename TT>
ostream& operator<<(ostream& s, pair<T, TT> t) {
  return s << "(" << t.first << "," << t.second << ")";
}
template <typename T>
ostream& operator<<(ostream& s, vector<T> t) {
  for (int i = 0; i < ((int)((t).size())); i++) s << t[i] << " ";
  return s;
}
const string numbers = "0123456789ABCDEFGHI";
string baseConv(int n, int b) {
  int currQ = n;
  string str, output;
  while (true) {
    str += numbers[currQ % b];
    currQ /= b;
    if (currQ == 0) break;
  }
  for (int i = str.size() - 1; i >= 0; i--) {
    output += str[i];
  }
  return output;
}
void solveQues() {
  int t;
  cin >> t;
  int flag = 1;
  vector<int> nums(t);
  for (int i = 0; i < t; i++) {
    cin >> nums[i];
    if (nums[i] % 2 == 0) {
      cout << nums[i] / 2 << endl;
    } else {
      cout << (flag + nums[i]) / 2 << endl;
      flag *= -1;
    }
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  ;
  cin.tie(NULL);
  cout.tie(NULL);
  solveQues();
  return 0;
}