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
n = int(input())
even = []
odd = []
sumEven = 0
sumOdd = 0
ans = []
for i in range(n):
    t = int(input())
    if abs(t) % 2 == 0:
        even.append(i)
        sumEven += t//2
    else:
        odd.append(i)
        sumOdd += t//2
    ans.append(t//2)

# d = sumEven - sumOdd


# if d <= 0:
#     k = 0
#     for i in range(len(odd)):
#         if k >= d:
#             break
#         if ans[odd[i]] < 0:
#             ans[odd[i]] += 1
#             k += 1
# else:
#     for i in range(d):
#         ans[odd[i]] += 1
d = sumEven + sumOdd
k = 0
for i in odd:
    if k >= abs(d):
        break
    ans[i] += 1
    k += 1


for i in ans:
    print(i)