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;
long long int const mod = 1000000007;
long long int countDivisors(long long int n) {
  long long int cnt = 0;
  for (long long int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i)
        cnt++;
      else
        cnt = cnt + 2;
    }
  }
  return cnt;
}
long long int power(long long int x, long long int y) {
  long long int temp;
  if (y == 0) return 1;
  temp = power(x, y / 2) % mod;
  if (y % 2 == 0)
    return ((temp) * (temp)) % mod;
  else
    return (((x % mod)) * ((temp * temp) % mod)) % mod;
}
long long int gcd1(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd1(b % a, a);
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd1(a, b);
}
long long int arr[200009];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n;
  cin >> n;
  long long int sum = 0;
  int flag = 0;
  for (int i = 0; i < (int)n; i++) {
    cin >> arr[i];
    if (arr[i] % 2 == 0) {
      sum += arr[i] / 2;
    }
  }
  for (int i = 0; i < (int)n; i++) {
    if (arr[i] % 2 == 0) {
      cout << arr[i] / 2 << "\n";
    } else {
      if (flag) {
        if (arr[i] >= 0) {
          cout << arr[i] / 2 << "\n";
        } else {
          cout << (arr[i] - 1) / 2 << "\n";
        }
        flag = 0;
      } else {
        if (arr[i] >= 0) {
          cout << (arr[i] + 1) / 2 << "\n";
        } else {
          cout << (arr[i] / 2) << "\n";
        }
        flag = 1;
      }
    }
  }
}