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 MAX_N = 200000;
const int MAX_K = 200000;
struct Elm {
  long long v;
  int s, i;
  Elm() {}
  Elm(long long _v, int _s, int _i) : v(_v), s(_s), i(_i) {}
  bool operator<(const Elm &e) const {
    return v < e.v || (v == e.v && i < e.i);
  }
  void print() const { printf("Elm(%lld,%d,%d)\n", v, s, i); }
};
Elm es[MAX_N];
bool cmpei(const Elm &a, const Elm &b) { return a.i < b.i; }
int main() {
  int n, k, x;
  scanf("%d%d%d", &n, &k, &x);
  int ng = 0;
  for (int i = 0; i < n; i++) {
    int ai;
    scanf("%d", &ai);
    es[i] = Elm(abs(ai), (ai < 0) ? -1 : 1, i);
    if (ai < 0) ng ^= 1;
  }
  set<Elm> eset(es, es + n);
  while (k--) {
    Elm e = *eset.begin();
    eset.erase(eset.begin());
    if (!ng) {
      e.v -= x;
      if (e.v <= 0) {
        e.v = abs(e.v), e.s *= -1;
        ng = 1;
      }
    } else {
      e.v += x;
    }
    eset.insert(e);
  }
  copy(eset.begin(), eset.end(), es);
  sort(es, es + n, cmpei);
  for (int i = 0; i < n; i++) {
    printf("%lld", es[i].v * es[i].s);
    putchar(i + 1 < n ? ' ' : '\n');
  }
  return 0;
}
