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 get_num(char ch) {
  if (ch == '-')
    return 0;
  else if (ch >= 'a' && ch <= 'z') {
    return 1 + (ch - 'a');
  } else if (ch >= 'A' && ch <= 'Z') {
    return 27 + (ch - 'A');
  } else if (ch >= '0' && ch <= '9') {
    return 53 + (ch - '0');
  }
}
inline long long pw(long long x, long long y) {
  if (y == 0) return 1;
  long long p = pw(x, y / 2);
  p *= p;
  if (y & 1) p *= x;
  return p;
}
inline int GA(long long n, vector<long long>& vec) {
  for (long long i = 0; i < n; ++i) {
    long long x;
    cin >> x;
    vec.push_back(x);
  }
  return 0;
}
inline int GAEG(long long m, vector<long long> vec[]) {
  for (long long i = 0; i < m; ++i) {
    long long x, y;
    cin >> x >> y;
    x--, y--;
    vec[x].push_back(y);
    vec[y].push_back(x);
  }
  return 0;
}
vector<long long> v;
bool vis[100000];
int main() {
  long long n;
  cin >> n;
  long long ss = 0;
  for (long long i = 0; i < n; ++i) {
    long long x;
    cin >> x;
    long long p = x / 2;
    if (x <= 0 && x % 2) p--;
    v.push_back(p);
    if ((x % 2)) vis[i] = true;
    ss += p;
  }
  if (!ss) {
    for (auto i : v) cout << i << endl;
  } else {
    for (long long i = 0; i < v.size(); ++i) {
      if (!ss) break;
      if (vis[i]) {
        ss++;
        v[i]++;
      }
    }
    for (auto i : v) cout << i << endl;
  }
  return 0;
}