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;
int n, k;
long long x;
long long ans[200005];
set<pair<long long, int> > ne, p;
pair<long long, int> aa;
vector<int> zeros;
int main() {
  cin >> n >> k >> x;
  for (int i = 0; i < n; i++) {
    long long aux;
    cin >> aux;
    if (aux == 0) zeros.push_back(i);
    if (aux < 0)
      ne.insert(make_pair(-aux, i));
    else
      p.insert(make_pair(aux, i));
  }
  for (; k > 0; k--) {
    if (zeros.size() > 0) {
      if (ne.size() % 2) {
        aa = make_pair(0LL, zeros[zeros.size() - 1]);
        p.erase(aa);
        zeros.pop_back();
        aa.first += x;
        p.insert(aa);
      } else {
        aa = make_pair(0LL, zeros[zeros.size() - 1]);
        p.erase(aa);
        zeros.pop_back();
        aa.first += x;
        ne.insert(aa);
      }
    } else {
      if (ne.size() % 2) {
        if (p.size() == 0) {
          aa = *ne.begin();
          ne.erase(aa);
          aa.first += x;
          ne.insert(aa);
        } else if ((*p.begin()).first < (*ne.begin()).first) {
          aa = *p.begin();
          p.erase(aa);
          aa.first += x;
          p.insert(aa);
        } else {
          aa = *ne.begin();
          ne.erase(aa);
          aa.first += x;
          ne.insert(aa);
        }
      } else {
        if (p.size() == 0 || ne.size() == 0) {
          if (p.size() == 0) {
            aa = *ne.begin();
            ne.erase(aa);
            aa.first -= x;
            if (aa.first <= 0) {
              if (aa.first == 0) zeros.push_back(aa.second);
              aa.first = -aa.first;
              p.insert(aa);
            } else
              ne.insert(aa);
          } else {
            aa = *p.begin();
            p.erase(aa);
            aa.first -= x;
            if (aa.first < 0) {
              aa.first = -aa.first;
              ne.insert(aa);
            } else {
              if (aa.first == 0) zeros.push_back(aa.second);
              p.insert(aa);
            }
          }
        } else if ((*p.begin()).first < (*ne.begin()).first) {
          aa = *p.begin();
          p.erase(aa);
          aa.first -= x;
          if (aa.first >= 0) {
            if (aa.first == 0) zeros.push_back(aa.second);
            p.insert(aa);
          } else {
            aa.first = -aa.first;
            ne.insert(aa);
          }
        } else {
          aa = *ne.begin();
          ne.erase(aa);
          aa.first -= x;
          if (aa.first <= 0) {
            if (aa.first == 0) zeros.push_back(aa.second);
            aa.first = -aa.first;
            p.insert(aa);
          } else
            ne.insert(aa);
        }
      }
    }
  }
  for (set<pair<long long, int> >::iterator it = p.begin(); it != p.end(); it++)
    ans[(*it).second] = (*it).first;
  for (set<pair<long long, int> >::iterator it = ne.begin(); it != ne.end();
       it++)
    ans[(*it).second] = -(*it).first;
  cout << ans[0];
  if (n == 1) {
    cout << endl;
    return 0;
  }
  for (int i = 1; i < n; i++) cout << " " << ans[i];
  cout << endl;
}
