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;
long long a[1000005] = {};
int main() {
  long long n, k, x;
  cin >> n >> k >> x;
  long long p = 0, e = 0, z = 0;
  vector<pair<long long, int> > ds;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == 0)
      z++;
    else if (a[i] > 0)
      p++;
    else
      e++;
    ds.push_back(make_pair(abs(a[i]), i));
  }
  sort(ds.begin(), ds.end());
  if (z > k) {
    a[0] -= k * x;
    k = 0;
  } else if (z <= k) {
    for (int i = 0; k >= 0 && i < n && z >= 0; i++) {
      if (ds[i].first == 0 && e % 2 == 0) {
        ds[i].first += x;
        int p = ds[i].second;
        a[p] -= x;
        e++;
        z--;
        k--;
      } else if (ds[i].first == 0 && e % 2 != 0) {
        ds[i].first += x;
        int p = ds[i].second;
        a[p] += x;
        z--;
        k--;
      }
    }
    sort(ds.begin(), ds.end());
  }
  if (k > 0) {
    long long p = ds[0].first;
    if (e % 2 != 0) {
      for (int i = 0; i < n && k > 0; i++) {
        if (ds[i].first <= p + x) {
          int lo = ds[i].second;
          if (a[lo] < 0)
            a[lo] -= x;
          else
            a[lo] += x;
          ds[i].first += x;
          k--;
          if (k <= 0) break;
        } else {
          i = -1;
          sort(ds.begin(), ds.end());
          p = ds[0].first;
        }
        if (i == n - 1 && k > 0) i = -1;
      }
    } else {
      int lo = ds[0].second, mo, po;
      if (a[lo] < 0) {
        mo = (-1 * a[lo]) / x;
        if (mo * x + a[lo] <= 0) mo++;
      } else if (a[lo] >= 0) {
        mo = (a[lo]) / x;
        if (-1 * mo * x + a[lo] >= 0) mo++;
      }
      if (k >= mo) {
        if (e > 0)
          e--;
        else
          e++;
        if (a[lo] < 0) {
          ds[0].first = abs(mo * x + a[lo]);
          a[lo] += mo * x;
        } else {
          ds[0].first = abs(-1 * mo * x + a[lo]);
          a[lo] -= mo * x;
        }
        k -= mo;
        sort(ds.begin(), ds.end());
        if (k > 0) {
          long long p = ds[0].first;
          for (int i = 0; i < n && k > 0; i++) {
            if (ds[i].first <= p + x) {
              int lo = ds[i].second;
              if (a[lo] < 0)
                a[lo] -= x;
              else
                a[lo] += x;
              ds[i].first += x;
              k--;
            } else {
              i = -1;
              sort(ds.begin(), ds.end());
              p = ds[0].first;
            }
            if (i == n - 1 && k > 0) i = -1;
          }
        }
      } else {
        if (a[lo] < 0)
          a[lo] += k * x;
        else
          a[lo] -= k * x;
      }
    }
  }
  for (int i = 0; i < n; i++) cout << a[i] << " ";
}
