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 long long isqrt(long long k) {
  long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline long long icbrt(long long k) {
  long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline long long mnz(long long& a, long long b) { return a = (a > b ? b : a); }
inline long long mxz(long long& a, long long b) { return a = (a < b ? b : a); }
inline string toString(long long n) {
  stringstream ss;
  ss << n;
  return ss.str();
}
double const eps = 1e-6;
long long const Base = 1e9 + 7, oo = 1e17, MAXN = 1e6;
long long A[MAXN + 5];
long long Solves() {
  long long n, m, k, cnt = 0, ans = 0, x, y, q, c, sum = 0, v, t;
  cin >> n;
  long long flag1 = 0, flag2 = 0;
  for (long long i = 1; i <= n; i++) {
    cin >> A[i];
    if (abs(A[i]) % 2 == 0)
      A[i] /= 2;
    else {
      if (A[i] > 0) {
        if (flag1)
          A[i] = A[i] / 2 + 1;
        else
          A[i] /= 2;
        flag1 ^= 1;
      } else {
        if (flag2)
          A[i] = A[i] / 2 - 1;
        else
          A[i] /= 2;
        flag2 ^= 1;
      }
    }
    cout << A[i] << endl;
  }
  return 0;
}
int main() {
  if (fopen("locin.txt", "r")) freopen("locin.txt", "r", stdin);
  long long JUDGE_ONLINE = 1;
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long test = 1;
  for (long long T = 1; T <= test; T++) {
    Solves();
  }
  if (!JUDGE_ONLINE)
    cout << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
}