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 INF = 0x3f3f3f3f;
const double PI = acos(-1.0);
const double E = exp(1);
const int N = 2e5 + 5;
long long n, k, x;
long long a[N];
int ct = 0, zero = 0;
int imen = 0;
bool sozero() { return zero > k; }
bool posi() {
  if ((ct % 2)) return false;
  if (zero) return false;
  int i = imen;
  return abs(a[i]) - k * x >= 0;
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin >> n >> k >> x;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (abs(a[i]) < abs(a[imen])) imen = i;
    if (a[i] == 0) zero++;
    if (a[i] < 0) ct++;
  }
  if (sozero()) {
    cout << a[0] + k * x << " ";
    for (int i = 1; i < n; i++) cout << a[i] << " ";
    cout << endl;
    return 0;
  }
  if (posi()) {
    for (int i = 0; i < n; i++) {
      if (i == imen) {
        if (a[i] > 0) {
          cout << a[i] - k * x << " ";
        } else {
          cout << a[i] + k * x << " ";
        }
      } else {
        cout << a[i] << " ";
      }
    }
    cout << endl;
    return 0;
  }
  set<pair<long long, int> > s;
  for (int i = 0; i < n; i++) {
    if (!a[i]) {
      if (!(ct % 2)) {
        a[i] -= x;
        k--;
        ct++;
      } else {
        a[i] += x;
        k--;
      }
    }
  }
  if (!(ct % 2)) {
    int i = imen;
    if (a[i] > 0) {
      while (a[i] >= 0) {
        a[i] -= x;
        k--;
      }
    } else {
      while (a[i] <= 0) {
        a[i] += x;
        k--;
      }
    }
  }
  for (int i = 0; i < n; i++) s.insert(make_pair(abs(a[i]), i));
  while (k--) {
    int i = s.begin()->second;
    s.erase(s.begin());
    if (a[i] < 0) {
      a[i] -= x;
    } else {
      a[i] += x;
    }
    s.insert(make_pair(abs(a[i]), i));
  }
  for (int i = 0; i < n; i++) cout << a[i] << " ";
  cout << endl;
  return 0;
}
