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,x = map(int, raw_input().split())
a = map(int, raw_input().split())
q = []
checker = 1
for i in xrange(n):
  heapq.heappush(q, (abs(a[i]), i))
  if a[i] < 0: checker *= -1
while k > 0:
  nxt = heapq.heappop(q)
  if checker > 0:
    if a[nxt[1]] >= 0:
      a[nxt[1]] -= x
      if a[nxt[1]] < 0: checker *= -1
    else:
      a[nxt[1]] += x
      if a[nxt[1]] >= 0: checker *= -1
  else:
    if a[nxt[1]] >= 0:
      a[nxt[1]] += x
    else:
      a[nxt[1]] -= x
  heapq.heappush(q, (abs(a[nxt[1]]), nxt[1]))
  k -= 1
print " ".join(map(str, a))
