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>
const long long mod = 1000000007;
const long long Inf = 1e9;
using namespace std;
int main() {
  int n, k, x;
  scanf("%d %d %d", &n, &k, &x);
  vector<pair<long long, long long>> a(n);
  vector<long long> ans(n), abss(n, 1);
  bool flag = false;
  for (int(i) = 0; (i) < n; (i)++) {
    long long tmp;
    scanf("%I64d", &tmp);
    if (tmp < 0) {
      flag ^= 1;
      abss[i] = -1;
    }
    a[i] = make_pair(abs(tmp), i);
  }
  sort((a).begin(), (a).end());
  if (!flag) {
    int q = a[0].first / x;
    if (q + 1 <= k) {
      if (abss[a[0].second] == 1)
        abss[a[0].second] = -1;
      else
        abss[a[0].second] = 1;
      a[0].first = abs(a[0].first - x * (q + 1));
      flag = true;
      k -= q + 1;
    }
  }
  if (!flag) {
    a[0].first -= (long long)x * k;
    for (int(i) = 0; (i) < n; (i)++)
      ans[a[i].second] = a[i].first * abss[a[i].second];
    for (int(i) = 0; (i) < n; (i)++) cout << ans[i] << " ";
    return 0;
  }
  int i = 0;
  map<pair<long long, long long>, long long> b;
  for (int(i) = 0; (i) < n; (i)++) b[make_pair(a[i].first, a[i].second)]++;
  auto it = b.begin();
  while (k > 0) {
    if (b[make_pair(it->first.first, it->first.second)] > 0) {
      b[make_pair(it->first.first, it->first.second)]--;
      b[make_pair(it->first.first + x, it->first.second)]++;
      k--;
    } else
      it++;
  }
  for (auto(it) = (b).begin(); (it) != (b).end(); (it)++) {
    if (b[make_pair(it->first.first, it->first.second)] > 0) {
      ans[it->first.second] = it->first.first * abss[it->first.second];
    }
  }
  for (int(i) = 0; (i) < n; (i)++) printf("%I64d ", ans[i]);
  return 0;
}
