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 main() {
  int n, k;
  long long x, val;
  cin >> n >> k >> x;
  vector<pair<long long, int>> values(n);
  priority_queue<pair<long long, int>> pq;
  int minN = -1, minP = -1, numNeg = 0;
  for (int i = 0; i < n; ++i) {
    cin >> val;
    values[i] = (make_pair(val, i));
    if (values[i].first < 0) {
      ++numNeg;
      if (minN == -1)
        minN = i;
      else if (values[i].first > values[minN].first)
        minN = i;
    }
    if (values[i].first >= 0) {
      if (minP == -1)
        minP = i;
      else if (values[i].first < values[minP].first)
        minP = i;
    }
  }
  if (numNeg % 2 == 0) {
    if (minP == -1) {
      while (k && values[minN].first < 1) values[minN].first += x, --k;
    } else if (minN == -1) {
      while (k && values[minP].first > -1) values[minP].first -= x, --k;
    } else {
      int p = values[minP].first, neg = values[minN].first;
      while (k && p >= 0 && neg <= 0) {
        --k;
        neg += x;
        p -= x;
      }
      if (p < 0 && abs(p) < abs(neg)) {
        values[minN].first = neg;
      } else {
        values[minP].first = p;
      }
    }
  }
  vector<long long> minu(n);
  for (int i = 0; i < n; ++i) {
    if (values[i].first < 0)
      minu[i] = -1;
    else
      minu[i] = 1;
    values[i].first = -abs(values[i].first);
    pq.push(values[i]);
  }
  while (k) {
    pair<long long, int> tmp;
    tmp = pq.top();
    pq.pop();
    tmp.first -= x;
    pq.push(tmp);
    --k;
  }
  vector<long long> ans(n);
  for (int i = 0; i < n; ++i) {
    pair<long long, int> tmp = pq.top();
    pq.pop();
    ans[tmp.second] = -tmp.first;
  }
  for (int i = 0; i < n; ++i) cout << ans[i] * minu[i] << " ";
  cout << endl;
  return 0;
}
