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 = 200000;
const long long INFLL = 1000000000000000000LL;
long long v[1 + MAXN];
long long Abs(long long x) {
  if (x < 0) return -x;
  return x;
}
class Compare {
 public:
  bool operator()(const int &x, const int &y) { return Abs(v[x]) > Abs(v[y]); }
};
priority_queue<int, vector<int>, Compare> Heap;
int main() {
  int n, k, sign = 1, zeros = 0;
  long long x;
  scanf("%d%d%I64d", &n, &k, &x);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &v[i]);
    if (v[i] < 0) sign = -sign;
    if (v[i] == 0) zeros++;
  }
  if (zeros) {
    if (zeros > k) {
      for (int i = 1; i <= k; i++) v[i] += x;
      for (int i = 1; i <= n; i++) printf("%I64d ", v[i]);
      return 0;
    }
    k -= zeros;
    bool first = true;
    for (int i = 1; i <= n; i++)
      if (v[i] == 0)
        if (!first)
          v[i] += x;
        else {
          first = false;
          if (sign == 1)
            v[i] -= x;
          else
            v[i] += x;
        }
  } else if (sign == 1) {
    long long smallest = INFLL;
    int where;
    for (int i = 1; i <= n; i++)
      if (Abs(v[i]) < smallest) {
        smallest = Abs(v[i]);
        where = i;
      }
    long long need = smallest / x + 1;
    if (need >= k) {
      if (v[where] > 0)
        v[where] = v[where] - x * k;
      else
        v[where] = v[where] + x * k;
      for (int i = 1; i <= n; i++) printf("%I64d ", v[i]);
      return 0;
    }
    k -= need;
    if (v[where] > 0)
      v[where] = v[where] - x * need;
    else
      v[where] = v[where] + x * need;
  }
  for (int i = 1; i <= n; i++) Heap.push(i);
  for (int i = 1; i <= k; i++) {
    int where = Heap.top();
    Heap.pop();
    if (v[where] < 0)
      v[where] -= x;
    else
      v[where] += x;
    Heap.push(where);
  }
  for (int i = 1; i <= n; i++) printf("%I64d ", v[i]);
  return 0;
}
