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 Number {
  long long num;
  int pos;
  bool operator>(const Number &rhs) const { return abs(num) > abs(rhs.num); }
};
int n, c, ans, k, x;
long long a[210000];
bool flag = 0;
priority_queue<Number, vector<Number>, greater<Number> > Q;
int main() {
  scanf("%d%d%d", &n, &k, &x);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
    Q.push((Number){a[i], i});
    if (a[i] < 0) flag ^= 1;
  }
  int now = 1;
  Number b;
  if (!flag) {
    b = Q.top();
    Q.pop();
    if (b.num >= 0) {
      for (now = 1; now <= k && b.num >= 0; now++) {
        b.num -= x;
      }
    } else {
      for (now = 1; now <= k && b.num <= 0; now++) {
        b.num += x;
      }
    }
    Q.push(b);
  }
  for (; now <= k; now++) {
    b = Q.top();
    Q.pop();
    if (b.num >= 0)
      b.num += x;
    else
      b.num -= x;
    Q.push(b);
  }
  while (!Q.empty()) {
    b = Q.top();
    Q.pop();
    a[b.pos] = b.num;
  }
  for (int i = 1; i < n; i++) {
    printf("%I64d ", a[i]);
  }
  printf("%I64d\n", a[n]);
}
