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 maxn = 2e5 + 5;
struct node {
  long long a;
  int b;
  int id;
  bool operator<(const node& aa) const {
    if (aa.a == a) return b > aa.b;
    return a > aa.a;
  }
};
long long b[maxn];
int main(void) {
  int n;
  long long k, x;
  scanf("%d%I64d%I64d", &n, &k, &x);
  long long u;
  int cnt = 0;
  priority_queue<node> q;
  for (int i = 0; i < n; ++i) {
    scanf("%I64d", &u);
    q.push((node){u >= 0 ? u : -u, u >= 0 ? 1 : -1, i});
    if (u < 0) cnt++;
  }
  if (cnt & 1) {
    while (k) {
      node tmp = q.top();
      q.pop();
      tmp.a += x;
      q.push((node){tmp.a, tmp.b, tmp.id});
      k--;
    }
  } else {
    bool flag = false;
    while (k) {
      node t = q.top();
      q.pop();
      int tmp = t.a / x;
      if (!flag) {
        if (tmp < k) {
          k -= tmp + 1;
          t.a -= (tmp + 1) * x;
          t.b = -t.b;
          q.push((node){t.a >= 0 ? t.a : -t.a, t.b, t.id});
        } else {
          t.a -= k * x;
          q.push((node){t.a >= 0 ? t.a : -t.a, t.b, t.id});
          break;
        }
        flag = true;
      } else {
        t.a += x;
        k--;
        q.push((node){t.a >= 0 ? t.a : -t.a, t.b, t.id});
      }
    }
  }
  while (!q.empty()) {
    node t = q.top();
    q.pop();
    b[t.id] = t.a * t.b;
  }
  for (int i = 0; i < n; ++i) {
    printf("%I64d ", b[i]);
  }
  puts("");
  return 0;
}
