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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
inline long long rand(long long x, long long y) {
  return (rng() % (y + 1 - x)) + x;
}
string to_string(char c) {
  string second(1, c);
  return second;
}
template <typename T>
inline T gcd(T a, T b) {
  return a == 0 ? b : gcd(b % a, a);
}
long long int n;
long long int A[(300006)];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (long long int i = (0); i <= (long long int)(n - 1); ++i) cin >> A[i];
  long long int up = 0;
  for (long long int i = (0); i <= (long long int)(n - 1); ++i) {
    if (A[i] & 1) {
      if (up & 1) {
        if (A[i] > 0)
          cout << (A[i] + 1) / 2 << '\n';
        else
          cout << A[i] / 2 << '\n';
        ++up;
      } else {
        if (A[i] > 0)
          cout << A[i] / 2 << '\n';
        else
          cout << (A[i] - 1) / 2 << '\n';
        ++up;
      }
    } else {
      cout << A[i] / 2 << '\n';
    }
  }
}