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:
import math
n=int(input())
oddp=[]
oddn=[]
sumarr=[]
ceil=[]
floor=[]
for _ in range(n):
	#n=int(input())
	#for n1 in range(n):
	a=int(input())
	sumarr.append(a)
	ceil.append(math.ceil(a/2))
	floor.append(math.floor(a/2))
	if a&1:
		if a <= 0:
			oddn.append(a)
		else:
			oddp.append(a)
# print(*floor)
# print(*ceil)
# print(*oddp)
# print(*oddn)
count1=abs(sum(floor))
count2=abs(sum(ceil))
if count1==0:
	for u in floor:
		print(u)
elif count2==0:
	for u in ceil:
		print(u)
else:
	if count1 > count2:
		for i in range(0,n):
			if count2>0:
				if math.ceil(sumarr[i]/2)!=math.floor(sumarr[i]/2):
					sumarr[i]=math.floor(sumarr[i]/2)
					count2=count2-1
				else:
					sumarr[i]=sumarr[i]//2 
			else:
				if math.ceil(sumarr[i]/2)!=math.floor(sumarr[i]/2):
					sumarr[i]=math.ceil(sumarr[i]/2)
				else:
					sumarr[i]=sumarr[i]//2
	else:
		for i in range(0,n):
			if count1>0:
				if math.ceil(sumarr[i]/2)!=math.floor(sumarr[i]/2):
					sumarr[i]=math.ceil(sumarr[i]/2)
					count1=count1-1
				else:
					sumarr[i]=sumarr[i]//2 
			else:
				if math.ceil(sumarr[i]/2)!=math.floor(sumarr[i]/2):
					sumarr[i]=math.floor(sumarr[i]/2)
				else:
					sumarr[i]=sumarr[i]//2
	for s in sumarr:
		print(s)