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 
import heapq

n,k,v = map(int, input().split())
num = list(map(int, input().split()))

sign = len([x for x in num if x<0])
minx,mini = min([[abs(x),i] for i,x in enumerate(num)])

if sign%2==0:
    if num[mini]<0:
        c = min(minx//v+1, k)
        k -= c
        num[mini] += v*c
    elif num[mini]>=0:
        c = min(minx//v+1, k)
        k -= c
        num[mini] -= v*c


heap = []
heapq.heapify(heap)
for i,x in enumerate(num):
    heapq.heappush(heap, [abs(x), i])

while k:
    absx,curi = heapq.heappop(heap)
    #print(curi, num[curi])
    if num[curi]>=0:
        num[curi] += v
    else:
        num[curi] -= v
    heapq.heappush(heap, [abs(num[curi]),curi])
    k -= 1
    
print(' '.join([str(x) for x in num]))
    