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;
const long long int inf = 1e18;
const long long int mod = 1e9 + 7;
inline long long int add(long long int x, long long int y) {
  x += y;
  if (x >= mod) x -= mod;
  return x;
}
inline long long int sub(long long int x, long long int y) {
  x -= y;
  if (x < 0) x += mod;
  return x;
}
inline long long int mul(long long int x, long long int y) {
  return ((x % mod) * (y % mod)) % mod;
}
inline long long int gcd(long long int x, long long int y) {
  return x == 0 ? y : gcd(y % x, x);
}
inline long long int power(long long int a, long long int b) {
  long long int x = 1;
  while (b) {
    if (b & 1) x = mul(x, a);
    a = mul(a, a);
    b >>= 1;
  }
  return x;
}
inline long long int inv(long long int a) { return power(a, mod - 2); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, flag = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    if (x & 1) {
      x = x >> 1;
      if (flag) x++;
      flag = 1 - flag;
      cout << x << "\n";
    } else
      cout << x / 2 << "\n";
  }
  return 0;
}