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 min(long long int a, long long int b) { return a < b ? a : b; }
long long int min(long long int a, long long int b, long long int c) {
  a = min(a, b);
  a = min(a, c);
  return a;
}
long long int max(long long int a, long long int b) { return a > b ? a : b; }
long long int max(long long int a, long long int b, long long int c) {
  a = max(a, b);
  a = max(a, c);
  return a;
}
long long int dsum(long long int n) {
  long long int ans = 0;
  while (n) {
    ans += (n % 10);
    n /= 10;
  }
  return ans;
}
long long int power(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long int Ceil(long long int a, long long int b) {
  long long int res = a / b;
  if (a % b != 0) res++;
  return res;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &v) {
  for (T &t : v) is >> t;
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (const T &t : v) {
    os << t << "\n";
  }
  os << '\n';
  return os;
}
long long int dxx[] = {1, 1, 0, -1, -1, -1, 0, 1};
long long int dyy[] = {0, 1, 1, 1, 0, -1, -1, -1};
long long int dx[] = {1, 0, -1, 0};
long long int dy[] = {0, 1, 0, -1};
struct pll {
  long long int x, y;
};
bool comp(pll a, pll b) {
  if (a.x == b.x)
    return a.y < b.y;
  else
    return a.x < b.x;
}
void solve(long long int the) {
  char ch;
  double aa, bb, cc, dd, xx;
  string s1, s2, s3, str;
  long long int i, j, k, a, b, c, d, n, m, l, r, x, y, z, low, high, mid,
      sum = 0, ans = 0, temp, t;
  cin >> n;
  vector<long long int> v(n);
  cin >> v;
  vector<long long int> vv(n);
  vector<double> v1;
  for (i = 0; i < n; i++) v1.push_back(v[i]);
  c = 0;
  for (i = 0; i < n; i++) {
    if (v[i] % 2 == 0) {
      vv[i] = v[i] / 2;
    }
  }
  a = 0;
  for (i = 0; i < n; i++) {
    if ((v[i] < 0) && (v[i] % 2 != 0)) {
      if (a == 0) {
        a = 1;
        vv[i] = floor(v1[i] / 2);
      } else {
        a = 0;
        vv[i] = ceil(v1[i] / 2);
      }
    }
  }
  a = 0;
  for (i = 0; i < n; i++) {
    if ((v[i] > 0) && (v[i] % 2 != 0)) {
      if (a == 0) {
        a = 1;
        vv[i] = ceil(v1[i] / 2);
      } else {
        a = 0;
        vv[i] = floor(v1[i] / 2);
      }
    }
  }
  cout << vv;
}
int main() {
  long long int t = 1, c = 1;
  while (t--) {
    solve(c);
    c++;
  }
  return 0;
}