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 PYTHON3):

Solution:
from bisect import *
from collections import *
from itertools import *
import functools
import sys
import math
from decimal import *
from copy import *
from heapq import *
from fractions import *
getcontext().prec = 30
MAX = sys.maxsize
MAXN = 1000010
MOD = 10**9+7
spf = [i for i in range(MAXN)]
def sieve():
    for i in range(2,MAXN,2):
        spf[i] = 2
    for i in range(3,int(MAXN**0.5)+1):
        if spf[i]==i:
            for j in range(i*i,MAXN,i):
                if spf[j]==j:
                    spf[j]=i
def fib(n,m):
    if n == 0:
        return [0, 1]
    else:
        a, b = fib(n // 2)
        c = ((a%m) * ((b%m) * 2 - (a%m)))%m
        d = ((a%m) * (a%m))%m + ((b)%m * (b)%m)%m
        if n % 2 == 0:
            return [c, d]
        else:
            return [d, c + d]

def charIN(x= ' '):
    return(sys.stdin.readline().strip().split(x))

def arrIN(x = ' '):
    return list(map(int,sys.stdin.readline().strip().split(x)))

def ncr(n,r):
    num=den=1
    for i in range(r):
        num = (num*(n-i))%MOD
        den = (den*(i+1))%MOD

    return (num*(pow(den,MOD-2,MOD)))%MOD

def flush():
    return sys.stdout.flush()


'''*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*'''

a = [int(input()) for _ in range(int(input()))]
o = 0
for i in a:
    if i%2:
        o+=1
assert(o%2==0)
x = o//2
f = 1
b = []
for i in a:
    if i%2:
        if f:
            b+=[i//2]
            x-=1
            if x==0:
                f^=1
        else:
            b+=[(i+1)//2]
    else:
        b+=[i//2]
for i in b:
    print(i)