721_D. Maxim and Array
Recently Maxim has found an array of n integers, needed by no one. He immediately come up with idea of changing it: he invented positive integer x and decided to add or subtract it from arbitrary array elements. Formally, by applying single operation Maxim chooses integer i (1 ≤ i ≤ n) and replaces the i-th element of array ai either with ai + x or with ai - x. Please note that the operation may be applied more than once to the same position.

Maxim is a curious minimalis, thus he wants to know what is the minimum value that the product of all array elements (i.e. <image>) can reach, if Maxim would apply no more than k operations to it. Please help him in that.

Input

The first line of the input contains three integers n, k and x (1 ≤ n, k ≤ 200 000, 1 ≤ x ≤ 109) — the number of elements in the array, the maximum number of operations and the number invented by Maxim, respectively.

The second line contains n integers a1, a2, ..., an (<image>) — the elements of the array found by Maxim.

Output

Print n integers b1, b2, ..., bn in the only line — the array elements after applying no more than k operations to the array. In particular, <image> should stay true for every 1 ≤ i ≤ n, but the product of all array elements should be minimum possible.

If there are multiple answers, print any of them.

Examples

Input

5 3 1
5 4 3 5 2


Output

5 4 3 5 -1 


Input

5 3 1
5 4 3 5 5


Output

5 4 0 5 5 


Input

5 3 1
5 4 4 5 5


Output

5 1 4 5 5 


Input

3 2 7
5 4 2


Output

5 11 -5 
```python
import heapq as hq
from math import ceil
n, k, x = [int(i) for i in input().strip().split(' ')]
arr = [int(i) for i in input().strip().split(' ')]
is_neg = False
for i in arr:
    if i < 0:
        is_neg = True if is_neg == False else False

narr = [[abs(i), pos, i < 0] for pos, i in enumerate(arr)]
hq.heapify(narr)
if is_neg:
    while k > 0:
        hq.heapreplace(narr, [narr[0][0]+x, narr[0][1], narr[0][2]])
        k -= 1
else:
    minode = hq.heappop(narr)
    mi = minode[0]
    kswitch = ceil(mi/x) #make the off number of negatives
    if kswitch > k:
        kswitch = k
    else:
        minode[2] = False if minode[2] == True else True
    k -= kswitch

    hq.heappush(narr, [abs(mi-kswitch*x), minode[1], minode[2]])
    while k > 0:
        hq.heapreplace(narr, [narr[0][0]+x, narr[0][1], narr[0][2]])
        k -= 1

narr = sorted(narr, key=lambda x:x[1])
arr = [str(i[0]*(-1 if i[2] else 1)) for i in narr]
print(" ".join(arr))
```