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>
using namespace std;
const int maxN = 200100;
const long long vc = 1e9 * 2 + 7;
int H[maxN];
long long A[maxN], D[maxN];
int nH, n, k, x;
int demA, demD, minA, minD;
void UpH(int i) {
  int T = H[i];
  while (1) {
    int j = i / 2;
    if (j < 1) break;
    if (A[H[j]] > A[T]) {
      H[i] = H[j];
      i = j;
    } else
      break;
  }
  H[i] = T;
}
void DownH(int i) {
  int T = H[i];
  while (1) {
    int j = i * 2;
    if (j > nH) break;
    if (j < nH && A[H[j]] > A[H[j + 1]]) j++;
    if (A[H[j]] < A[T]) {
      H[i] = H[j];
      i = j;
    } else
      break;
  }
  H[i] = T;
}
void xl2() {
  for (int i = 1; i <= n; i++)
    if (A[i] == 0) {
      if (k == 0) return;
      A[i] += x;
      k--;
    }
  nH = 0;
  for (int i = 1; i <= n; i++) {
    nH++;
    H[nH] = i;
    UpH(nH);
  }
  while (k > 0) {
    A[H[1]] += x;
    k--;
    DownH(1);
  }
}
void xl1() {
  if (A[minA] < A[minD]) {
    int t = A[minA] / x;
    if (A[minA] % x != 0) t++;
    if (t >= k) {
      A[minA] -= k * x;
      return;
    }
    A[minA] = -(A[minA] - t * x);
    D[minA] *= -1;
    k -= t;
  } else {
    int t = A[minD] / x + 1;
    if (t >= k) {
      A[minD] -= k * x;
      return;
    }
    A[minD] = -(A[minD] - t * x);
    D[minD] *= -1;
    k -= t;
  }
  xl2();
}
int main() {
  cin >> n >> k >> x;
  demD = 0, demA = 0, minA = 0, minD = 0;
  A[0] = vc;
  D[0] = 1;
  for (int i = 1; i <= n; i++) {
    cin >> A[i];
    if (A[i] >= 0) {
      D[i] = 1;
      demD++;
      if (A[i] < A[minD]) minD = i;
    } else {
      D[i] = -1;
      demA++;
      A[i] *= -1;
      if (A[i] < A[minA]) minA = i;
    }
  }
  if (demA % 2 == 0)
    xl1();
  else
    xl2();
  for (int i = 1; i <= n; i++) cout << A[i] * D[i] << " ";
}
