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 
#include <bits/stdc++.h>
int n, k, x;
long long a[200010], A[200010];
struct item {
  int id;
  long long val;
  bool operator<(const item& it) const { return val > it.val; }
};
std::priority_queue<item> Q;
int main() {
  std::cin >> n >> k >> x;
  int sign = 1;
  for (int i = 0; i < n; i++) {
    std::cin >> a[i];
    if (a[i] < 0)
      sign = -sign, A[i] = -a[i];
    else
      A[i] = a[i];
    if (a[i] == 0) sign = 0;
  }
  if (sign > 0) {
    int mp = 0, t = 0;
    for (int i = 1; i < n; i++)
      if (A[i] < A[mp]) mp = i;
    while (k > 0 && A[mp] >= 0) {
      k--;
      t++;
      A[mp] -= x;
    }
    if (a[mp] > 0)
      a[mp] -= 1ll * t * x;
    else
      a[mp] += 1ll * t * x;
    if (k == 0) {
      for (int i = 0; i < n; i++) std::cout << a[i] << ' ';
      return 0;
    }
    A[mp] = -A[mp];
  }
  if (sign == 0) {
    int zcnt = 0;
    for (int i = 0; i < n; i++)
      if (a[i] == 0) zcnt++;
    if (k < zcnt) {
      for (int i = 0; i < n; i++) std::cout << a[i] << ' ';
      return 0;
    } else {
      int lst;
      for (int i = 0; i < n; i++)
        if (a[i] == 0) a[i] = A[i] = x, lst = i;
      sign = 1;
      for (int i = 0; i < n; i++)
        if (a[i] < 0) sign = -sign;
      if (sign == 1) a[lst] = -x;
      k -= zcnt;
    }
  }
  for (int i = 0; i < n; i++) Q.push((item){i, A[i]});
  while (k--) {
    item it = Q.top();
    Q.pop();
    if (a[it.id] > 0)
      a[it.id] += x;
    else
      a[it.id] -= x;
    Q.push((item){it.id, it.val + x});
  }
  for (int i = 0; i < n; i++) std::cout << a[i] << ' ';
  return 0;
}
