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 a, b, c, d, x, y, z, p, q, l, r, k,
    t = 1, n, m, cnt, curr, prev, ndx, i, j, rem, cnt1, cnt2, tmp, nxt, sum,
    ans, mx, mn;
long long mypow(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = res * x;
    y = y >> 1;
    x = x * x;
  }
  return res;
}
long long mylog(long long n, long long b) {
  long long cnt = 0;
  while (n > 1) n /= b, cnt++;
  return cnt;
}
bool cmp(pair<long long, long long> p1, pair<long long, long long> p2) {
  return true;
}
pair<long long, long long> manacher(string s) {
  int n = s.size();
  vector<int> d1(n);
  for (int i = 0, l = 0, r = -1; i < n; i++) {
    int k = (i > r) ? 1 : min(d1[l + r - i], r - i + 1);
    while (0 <= i - k && i + k < n && s[i - k] == s[i + k]) {
      k++;
    }
    d1[i] = k--;
    if (i + k > r) {
      l = i - k;
      r = i + k;
    }
  }
  vector<int> d2(n);
  for (int i = 0, l = 0, r = -1; i < n; i++) {
    int k = (i > r) ? 0 : min(d2[l + r - i + 1], r - i + 1);
    while (0 <= i - k - 1 && i + k < n && s[i - k - 1] == s[i + k]) {
      k++;
    }
    d2[i] = k--;
    if (i + k > r) {
      l = i - k - 1;
      r = i + k;
    }
  }
  pair<long long, long long> ret = {0, 0};
  for (long long i = 0; i < n; i++) {
    if (i + 1 == d1[i] && (2 * (d1[i] - 1) + 1) > ret.first)
      ret.first = 2 * (d1[i] - 1) + 1;
    if (i == d2[i] && 2 * d2[i] > ret.first) ret.first = 2 * d2[i];
    if (i + d1[i] == n && (2 * (d1[i] - 1) + 1) > ret.second)
      ret.second = 2 * (d1[i] - 1) + 1;
    if (i + d2[i] == n && 2 * d2[i] > ret.second) ret.second = 2 * d2[i];
  }
  return (ret);
}
struct hash_pair {
  template <class T1, class T2>
  size_t operator()(const pair<T1, T2>& p) const {
    auto hash1 = hash<T1>{}(p.first);
    auto hash2 = hash<T2>{}(p.second);
    return hash1 ^ hash2;
  }
};
string s = "", s1 = "", s2 = "";
void solve() {
  cin >> n;
  vector<long long> v(n);
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
    rem += v[i] / 2;
  }
  if (rem > 0) {
    for (long long i = 0; i < n; i++) {
      if (rem > 0 && abs(v[i]) % 2 == 1 && v[i] < 0) {
        v[i] = (v[i] - 1) / 2;
        rem--;
      } else
        v[i] /= 2;
    }
  } else if (rem < 0) {
    rem = abs(rem);
    for (long long i = 0; i < n; i++) {
      if (rem > 0 && v[i] % 2 == 1 && v[i] > 0) {
        v[i] = (v[i] + 1) / 2;
        rem--;
      } else
        v[i] /= 2;
    }
  } else {
    for (long long i = 0; i < n; i++) v[i] /= 2;
  }
  for (long long i = 0; i < n; i++) cout << v[i] << endl;
  ;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  while (t--) solve();
}