human
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.
The above is tricky. Write me a correct solution in PYTHON3.
 What is a Python program to print any?
assistant
#Here is Python program to print any.

#    !/usr/bin/env python3
#    encoding: UTF-8
#    Modified: <16/Oct/2019 08:12:06 PM>

#    ✪ H4WK3yE乡
#    Mohd. Farhan Tahir
#    Indian Institute Of Information Technology (IIIT), Gwalior

import sys
import os
from io import IOBase, BytesIO

def main():
    from math import ceil, floor

    n = int(input())
    arr = [0] * n
    for i in range(n):
        arr[i] = int(input())
    f1, f2 = 0, 0
    ans = [0] * n
    for i in range(n):
        if arr[i] % 2 == 0:
            ans[i] = arr[i] // 2
        else:
            if arr[i] <= 0:
                if f1 == 0:
                    ans[i] = ceil(arr[i] / 2)
                else:
                    ans[i] = floor(arr[i] / 2)
                f1 = 1 - f1
            else:
                if f2 == 0:
                    ans[i] = floor(arr[i] / 2)
                else:
                    ans[i] = ceil(arr[i] / 2)
                f2 = 1 - f2
    for i in ans:
        print(i)

BUFSIZE = 8192

class FastIO(BytesIO):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0, 2), super(FastIO, self).write(s))[0])
        return s

    def read(self):
        while self._fill():
            pass
        return super(FastIO, self).read()

    def readline(self):
        while self.newlines == 0:
            s = self._fill()
            self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        py2 = round(0.5)
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
   ...