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())
x=0
a=[]
b=[]
for i in range(n):
    k=int(input())
    b.append(k)
    if (k%2==0):
        a.append(k//2)
    else:
        k=k/2
        if (abs(k-math.floor(k))>abs(k-math.ceil(k))):
            a.append(math.ceil(k))
        else:
            a.append(math.floor(k))
val=sum(a)
if (val!=0):
    for i in range(n):
        if (val==0):
            break
        if (b[i]%2!=0):
            if (val>0):
                if (math.ceil(b[i]/2)==a[i]):
                    a[i]=a[i]-1
                    val=val-1
            elif (val<0):
                if (math.floor(b[i]/2)==a[i]):
                    a[i]=a[i]+1
                    val=val+1
for i in range(n):
    print(a[i])