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;
inline int read() {
  int sum = 0;
  char c = getchar();
  bool f = 0;
  while (c < '0' || c > '9') {
    if (c == '-') f = 1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    sum = sum * 10 + c - '0';
    c = getchar();
  }
  if (f) return -sum;
  return sum;
}
struct ex {
  long long v, pos;
} G;
priority_queue<ex> q;
inline bool operator<(const ex &a, const ex &b) { return a.v > b.v; }
int n, m, x;
long long a[200005];
int num;
int main() {
  int i, j, p;
  n = read();
  m = read();
  x = read();
  for (i = 1; i <= n; i++) {
    a[i] = read();
    if (a[i] < 0) {
      num++;
      q.push((ex){-a[i], i});
    } else
      q.push((ex){a[i], i});
  }
  while (m--) {
    p = q.top().pos;
    q.pop();
    num -= (a[p] < 0);
    if (num & 1)
      a[p] += x;
    else
      a[p] -= x;
    num += (a[p] < 0);
    if (a[p] < 0)
      q.push((ex){-a[p], p});
    else
      q.push((ex){a[p], p});
  }
  for (i = 1; i <= n; i++) printf("%I64d ", a[i]);
  return 0;
}
