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;
void solve() {
  int N;
  cin >> N;
  vector<int> A(N), B(N);
  for (int i = 0; i < N; ++i) cin >> A[i];
  for (int i = 0; i < N; ++i) B[i] = A[i] / 2;
  int sum = accumulate(B.begin(), B.end(), 0);
  if (sum > 0) {
    for (int i = 0; i < N; ++i) {
      if (A[i] % 2 && A[i] < 0) {
        B[i]--;
        sum--;
      }
      if (sum == 0) break;
    }
  } else if (sum < 0) {
    for (int i = 0; i < N; ++i) {
      if (A[i] % 2 && A[i] > 0) {
        B[i]++;
        sum++;
      }
      if (sum == 0) break;
    }
  }
  for (int i = 0; i < N; ++i) cout << B[i] << '\n';
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  solve();
  return 0;
}