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 N = 200010;
int b[N], n, k, val, sign = 1;
long long a[N];
inline bool cmp(int x, int y) { return abs(a[x]) > abs(a[y]); }
inline int sgn(long long x) { return (x < 0ll ? -1 : 1); }
void read() {
  scanf("%d%d%d", &n, &k, &val);
  for (int i = 0; i < n; ++i) {
    scanf("%lld", a + i);
    b[i] = i;
    sign *= sgn(a[i]);
  }
}
void solve() {
  read();
  make_heap(b, b + n, cmp);
  while (k--) {
    int cursgn = sign * sgn(a[b[0]]);
    a[b[0]] -= cursgn * val;
    sign = cursgn * sgn(a[b[0]]);
    pop_heap(b, b + n, cmp);
    push_heap(b, b + n, cmp);
  }
  for (int i = 0; i < n; ++i) printf("%lld ", a[i]);
}
int main() {
  int t = 1;
  for (int i = 0; i < t; ++i) solve();
  return 0;
}
