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>
const int maxn = 2e5 + 2;
const long long mod = 1e9 + 7, inf = 1e18;
using namespace std;
struct data {
  long long legit, abs;
  int idx;
};
struct comp {
  bool operator()(const data& i, const data& j) { return i.abs < j.abs; }
};
long long a[maxn];
multiset<data, comp> s;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  long long x;
  cin >> x;
  int neg = 0, pos = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    neg += (a[i] < 0);
    pos += (a[i] > 0);
  }
  int z = n - neg - pos;
  bool pres = 0;
  if (z > k) {
    for (int i = 1; i <= n; i++) cout << a[i] << " \n"[i == n];
    return 0;
  }
  if (!z) {
    if (!(neg & 1)) {
      long long minn = inf;
      int ind;
      for (int i = 1; i <= n; i++) {
        if (max(a[i], -a[i]) < minn) {
          minn = max(a[i], -a[i]);
          ind = i;
        }
      }
      if (minn < k * x) {
        if (a[ind] > 0) {
          while (a[ind] >= 0) {
            a[ind] -= x;
            --k;
          }
        } else {
          while (a[ind] <= 0) {
            a[ind] += x;
            --k;
          }
        }
        pres = 1;
      }
    } else
      pres = 1;
    for (int i = 1; i <= n; i++) {
      data t = {a[i], max(a[i], -a[i]), i};
      s.emplace(t);
    }
  } else {
    int lesser, greater;
    if ((neg & 1) and (z & 1)) {
      lesser = z - 1;
      greater = 1;
    } else if (neg & 1) {
      lesser = z;
      greater = 0;
    } else if (!(neg & 1) and (z & 1)) {
      lesser = z;
      greater = 0;
    } else if (!(neg & 1)) {
      lesser = z - 1;
      greater = 1;
    }
    for (int i = 1; i <= n; i++) {
      if (a[i]) {
        data t = {a[i], max(a[i], -a[i]), i};
        s.emplace(t);
        continue;
      }
      if (lesser > 0) {
        --lesser;
        a[i] -= x;
      } else if (greater > 0) {
        --greater;
        a[i] += x;
      }
      data t = {a[i], max(a[i], -a[i]), i};
      s.emplace(t);
    }
    pres = 1;
    k -= z;
  }
  while (k > 0) {
    auto top = s.begin();
    data nxt = {top->legit, top->abs, top->idx};
    if (top->legit > 0 and pres) {
      nxt.legit += x;
    } else if (top->legit > 0) {
      nxt.legit -= x;
    } else if (top->legit < 0 and pres) {
      nxt.legit -= x;
    } else {
      nxt.legit += x;
    }
    nxt.abs = max(nxt.legit, -nxt.legit);
    s.erase(top);
    s.emplace(nxt);
    a[nxt.idx] = nxt.legit;
    k--;
  }
  for (int i = 1; i <= n; i++) cout << a[i] << " \n"[i == n];
  return 0;
}
