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;
typedef priority_queue<pair<long long, int>, vector<pair<long long, int>>,
                       greater<pair<long long, int>>>
    min_pair_heap;
void remove_zeros(min_pair_heap &heap, vector<bool> &sign, int &neg, int &zero,
                  int x, int &k) {
  if (zero > k) {
    return;
  }
  if (zero > 0) {
    auto top = heap.top();
    heap.pop();
    --zero;
    sign[top.second] = (neg % 2 == 0);
    if (neg % 2 == 0) {
      ++neg;
    }
    top.first = x;
    --k;
    heap.push(top);
  }
  while (zero > 0) {
    auto top = heap.top();
    heap.pop();
    --zero;
    top.first += x;
    --k;
    heap.push(top);
  }
}
void min_pos(min_pair_heap &heap, vector<bool> &sign, int &neg, int x, int &k) {
  if (neg % 2 == 1) {
    return;
  }
  auto top = heap.top();
  heap.pop();
  long long steps = min(static_cast<long long>(k), top.first / x);
  k -= steps;
  top.first -= steps * x;
  if (k > 0) {
    --k;
    top.first = x - top.first;
    sign[top.second] = !sign[top.second];
  }
  heap.push(top);
}
void max_neg(min_pair_heap &heap, int x, int &k) {
  for (; k > 0; --k) {
    auto top = heap.top();
    heap.pop();
    top.first += x;
    heap.push(top);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  int n, k, x;
  cin >> n >> k >> x;
  min_pair_heap heap;
  vector<bool> sign;
  int neg = 0, zero = 0;
  for (int i = 0; i < n; ++i) {
    long long a;
    cin >> a;
    sign.push_back(a < 0);
    heap.push(make_pair(abs(a), i));
    if (a == 0) {
      ++zero;
    } else if (a < 0) {
      ++neg;
    }
  }
  remove_zeros(heap, sign, neg, zero, x, k);
  min_pos(heap, sign, neg, x, k);
  max_neg(heap, x, k);
  vector<long long> answ(n);
  for (; !heap.empty(); heap.pop()) {
    auto top = heap.top();
    answ[top.second] = top.first * (sign[top.second] ? -1 : 1);
  }
  for (long long a : answ) {
    cout << a << ' ';
  }
  return 0;
}
