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 t, n, tot = 0;
long long a[200005], k, ans[200005];
struct ljh {
  long long x, y;
  int id;
  bool operator<(const ljh& a) const { return a.y < y; }
} e[200005];
bool cmp(ljh u, ljh v) { return u.y < v.y; }
priority_queue<ljh> q;
int main() {
  cin >> n >> t >> k;
  for (int i = 1; i <= n; i++) {
    cin >> e[i].x;
    e[i].y = abs(e[i].x);
    e[i].id = i;
    if (e[i].x < 0) tot++;
  }
  if (tot % 2 == 0) {
    sort(e + 1, e + n + 1, cmp);
    if (e[1].x < 0) {
      while (t && e[1].x <= 0) {
        t--;
        e[1].x += k;
        e[1].y = abs(e[1].x);
      }
    } else {
      while (t && e[1].x >= 0) {
        t--;
        e[1].x -= k;
        e[1].y = abs(e[1].x);
      }
    }
  }
  for (int i = 1; i <= n; i++) q.push(e[i]);
  while (t) {
    ljh now = q.top();
    q.pop();
    if (now.x >= 0)
      now.x += k;
    else
      now.x -= k;
    now.y = abs(now.x);
    q.push(now);
    t--;
  }
  while (!q.empty()) {
    ans[q.top().id] = q.top().x;
    q.pop();
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  return 0;
}
