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 nMax = 200010;
const long long inf = 1e18;
int n;
long long k, x;
vector<long long> a(nMax);
int minusCnt() {
  int cnt = 0;
  for (int i = 1; i <= n; i++)
    if (a[i] < 0) cnt++;
  return cnt;
}
bool chkminus() {
  if (minusCnt() % 2) return true;
  for (int i = 1; i <= n; i++) {
    if (abs(a[i]) < k * x) return true;
  }
  return false;
}
int findAbsmin() {
  int i, j = 1;
  long long mn = inf;
  for (i = 1; i <= n; i++) {
    if (mn > abs(a[i])) {
      mn = abs(a[i]);
      j = i;
    }
  }
  return j;
}
void positive() {
  int i, j = findAbsmin();
  if (a[j] < 0)
    a[j] += k * x;
  else
    a[j] -= k * x;
  for (i = 1; i <= n; i++) cout << a[i] << " ";
  puts("");
}
int main() {
  int i, j;
  cin >> n >> k >> x;
  for (i = 1; i <= n; i++) cin >> a[i];
  if (!chkminus()) {
    positive();
    return 0;
  }
  if (minusCnt() % 2 == 0) {
    i = findAbsmin();
    if (a[i] < 0)
      while (a[i] < 0) a[i] += x, k--;
    else
      while (a[i] >= 0) a[i] -= x, k--;
  }
  priority_queue<pair<long long, int> > pq;
  for (i = 1; i <= n; i++) pq.push(pair<long long, int>(-abs(a[i]), i));
  while (k--) {
    i = pq.top().second;
    pq.pop();
    if (a[i] < 0)
      a[i] -= x;
    else
      a[i] += x;
    pq.push(pair<long long, int>(-abs(a[i]), i));
  }
  for (i = 1; i <= n; i++) cout << a[i] << " ";
  puts("");
}
