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 bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
const int INF = 1 << 29;
inline int two(int n) { return 1 << n; }
inline int test(int n, int b) { return (n >> b) & 1; }
inline void set_bit(int& n, int b) { n |= two(b); }
inline void unset_bit(int& n, int b) { n &= ~two(b); }
inline int last_bit(int n) { return n & (-n); }
inline int ones(int n) {
  int res = 0;
  while (n && ++res) n -= n & (-n);
  return res;
}
using namespace std;
int arr[100000];
long long t, n, x;
int main() {
  cin >> t;
  for (int i = 0; i < t; i++) {
    cin >> arr[i];
  }
  int flag = 0;
  for (int i = 0; i < t; i++) {
    if (arr[i] % 2 == 0) {
      cout << arr[i] / 2 << "\n";
    } else {
      if (flag == 0)
        if (arr[i] > 0)
          cout << arr[i] / 2 + 1 << "\n";
        else
          cout << arr[i] / 2 << "\n";
      else {
        if (arr[i] > 0)
          cout << arr[i] / 2 << "\n";
        else
          cout << arr[i] / 2 - 1 << "\n";
      }
      flag = (flag + 1) % 2;
    }
  }
}