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;
struct erz {
  long long a;
  int b;
};
struct cmp {
  bool operator()(erz &p, erz &q) { return (p.a > q.a); }
};
priority_queue<erz, vector<erz>, cmp> Q;
erz mp(long long a, int b) {
  erz c;
  c.a = a, c.b = b;
  return (c);
}
long long wabs(long long x) {
  if (x < 0) x = -x;
  return (x);
}
int fh(long long x) {
  if (x >= 0)
    return 1;
  else
    return -1;
}
long long a[200005];
int main() {
  int i, n, m, f = 1;
  long long x;
  scanf("%d%d%I64d", &n, &m, &x);
  for (i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
    if (a[i] < 0) f *= -1;
    Q.push(mp(wabs(a[i]), i));
  }
  for (i = 1; i <= m; i++) {
    int c = Q.top().b;
    Q.pop();
    long long ac = a[c];
    if (f == 1) {
      if (a[c] >= 0)
        a[c] -= x;
      else
        a[c] += x;
    } else {
      if (a[c] < 0)
        a[c] -= x;
      else
        a[c] += x;
    }
    if (fh(ac) != fh(a[c])) f *= -1;
    Q.push(mp(wabs(a[c]), c));
  }
  printf("%I64d", a[1]);
  for (i = 2; i <= n; i++) printf(" %I64d", a[i]);
  printf("\n");
  return 0;
}
