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;
using ull = unsigned long long;
using ll = long long;
using ld = long double;
using D = double;
using ii = pair<int, int>;
using vi = vector<int>;
using vii = vector<ii>;
template <typename T>
T sqr(T a) {
  return a * a;
}
template <typename T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <typename T>
T sgn(T a) {
  return a > 0 ? 1 : (a < 0 ? -1 : 0);
}
template <typename T>
T abs(T a) {
  return a > 0 ? a : -a;
}
const int inf = 2e9;
const ld st = 0.000001;
const ld pi = acos((ld)-1);
const int MAXN = 100000;
const int mod = 1e9 + 7;
int main() {
  int n;
  cin >> n;
  vector<int> a;
  int x1 = 0, x2 = 0;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    if (x % 2 == 0) {
      cout << x / 2 << endl;
    } else {
      if (x < 0) {
        cout << (-1) * (abs(x) + x1) / 2 << endl;
        x1 ^= 1;
      } else {
        cout << (abs(x) + x2) / 2 << endl;
        x2 ^= 1;
      }
    }
  }
  return 0;
}