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 xo(long long int x, long long int y) {
  return (x | y) & (~x | ~y);
}
long long int bin_Expo(long long int x, long long int n) {
  if (x == 0) return 0;
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return bin_Expo(x * x, n / 2);
  else
    return x * bin_Expo(x * x, (n - 1) / 2);
}
long long int mod_Expo(long long int x, long long int n, long long int M) {
  if (x == 0) return 0;
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return mod_Expo((x * x) % M, n / 2, M);
  else
    return (x * mod_Expo((x * x) % M, (n - 1) / 2, M)) % M;
}
bool prime_check(long long int x) {
  bool prime = (x >= 2);
  for (long long int i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      prime = false;
      break;
    }
  }
  return prime;
}
long long int logg(long long int base, long long int x) {
  return (long long int)(log(x) / log(base));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    long long int n, k, j, q, y, h, i, cc;
    cin >> n;
    long long int a[n], b[n], c[n];
    for (long long int i = 0; i < n; i++) {
      cin >> a[i];
      if (a[i] % 2 != 0)
        b[i] = 1;
      else
        b[i] = 0;
      if (a[i] >= 0)
        c[i] = 1;
      else
        c[i] = 0;
    }
    cc = 0;
    for (long long int i = 0; i < n; i++) {
      a[i] = a[i] / 2;
      cc += a[i];
    }
    if (cc < 0) {
      for (long long int i = 0; i < n; i++) {
        if (b[i] == 1 && c[i] == 1) {
          a[i] = a[i] + 1;
          cc++;
        }
        if (cc == 0) break;
      }
    }
    if (cc > 0) {
      for (long long int i = 0; i < n; i++) {
        if (b[i] == 1 && c[i] == 0) {
          a[i] = a[i] - 1;
          cc--;
        }
        if (cc == 0) break;
      }
    }
    for (long long int i = 0; i < n; i++) cout << a[i] << '\n';
  }
  return 0;
}