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;
int main() {
  ios_base::sync_with_stdio(false);
  auto cmp = [](const pair<long long, long long> &p1,
                const pair<long long, long long> &p2) {
    return p1.second > p2.second;
  };
  long long n, k, x;
  scanf("%lld %lld %lld", &n, &k, &x);
  long long neg = 0;
  vector<pair<long long, long long> > numbers(n);
  vector<bool> signs(n);
  for (long long i = 0; i < n; i++) {
    long long d;
    scanf("%lld", &d);
    if (d < 0) ++neg;
    numbers[i] = make_pair(i, abs(d));
    signs[i] = (d >= 0);
  }
  bool increase = (neg % 2 != 0);
  make_heap(numbers.begin(), numbers.end(), cmp);
  while (k--) {
    pair<long long, long long> smallest = numbers.front();
    if (increase) {
      smallest.second += x;
    } else {
      smallest.second -= x;
      if (smallest.second <= 0) {
        increase = true;
        smallest.second *= -1;
        signs[smallest.first] = !signs[smallest.first];
      }
    }
    pop_heap(numbers.begin(), numbers.end(), cmp);
    numbers.pop_back();
    numbers.push_back(smallest);
    push_heap(numbers.begin(), numbers.end(), cmp);
  }
  sort(numbers.begin(), numbers.end(),
       [](const pair<long long, long long> &p1,
          const pair<long long, long long> &p2) {
         return p1.first <= p2.first;
       });
  for (long long i = 0; i < n; i++) {
    printf("%lld ", signs[i] ? numbers[i].second : -numbers[i].second);
  }
  printf("\n");
  return 0;
}
