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 n, k;
long long x;
long long rec[205000];
struct node {
  int id;
  long long num;
};
bool operator<(const node& a, const node& b) { return abs(a.num) > abs(b.num); }
int main() {
  scanf("%d%d%I64d", &n, &k, &x);
  int judge = 1;
  long long M = 1LL << 50;
  int tag = -1;
  for (int i = 1; i <= n; i++) {
    long long tmp;
    scanf("%I64d", &tmp);
    if (M > abs(tmp)) {
      M = abs(tmp);
      tag = i;
    }
    if (tmp != 0) judge = judge * (tmp / abs(tmp));
    rec[i] = tmp;
  }
  if (judge > 0) {
    if (rec[tag] == 0) {
      rec[tag] -= x;
      k--;
    } else if (rec[tag] > 0) {
      while (rec[tag] >= 0 && k > 0) {
        k--;
        rec[tag] -= x;
      }
    } else if (rec[tag] < 0) {
      while (rec[tag] <= 0 && k > 0) {
        k--;
        rec[tag] += x;
      }
    }
  }
  priority_queue<node> pq;
  for (int i = 1; i <= n; i++) {
    node tmp;
    tmp.id = i;
    tmp.num = rec[i];
    pq.push(tmp);
  }
  while (k) {
    k--;
    node tmp = pq.top();
    pq.pop();
    if (tmp.num >= 0) {
      tmp.num += x;
      rec[tmp.id] += x;
    } else {
      tmp.num -= x;
      rec[tmp.id] -= x;
    }
    pq.push(tmp);
  }
  for (int i = 1; i <= n; i++) printf("%I64d%c", rec[i], i == n ? '\n' : ' ');
}
