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;
template <typename T>
T next_int() {
  T x = 0, p = 1;
  char ch;
  do {
    ch = getchar();
  } while (ch <= ' ');
  if (ch == '-') {
    p = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + (ch - '0');
    ch = getchar();
  }
  return p * x;
}
string next_token() {
  char ch;
  string ans = "";
  do {
    ch = getchar();
  } while (ch <= ' ');
  while (ch > ' ') {
    ans += ch;
    ch = getchar();
  }
  return ans;
}
const long long INF = (long long)1e18;
const int INFINT = (int)1e9 + 227 + 1;
const int MAXN = (int)5000 + 2;
const int MOD = (int)1e9 + 7;
const long double EPS = 1e-9;
long long bin_pow(long long a, long long b) {
  if (!b) return 1;
  long long ans = bin_pow(a, b / 2);
  ans = ans * ans;
  if (b % 2) ans = ans * a;
  return ans;
}
int main() {
  int n = next_int<int>();
  int k = next_int<int>();
  long long x = next_int<int>();
  vector<pair<long long, pair<int, bool> > > a(n);
  int ko = 0;
  for (int i = 0; i < n; i++) {
    a[i] = make_pair(next_int<int>(), make_pair(i, 0));
    a[i].second.second = a[i].first < 0;
    ko += (a[i].first < 0);
  }
  sort(a.begin(), a.end());
  if (ko % 2 == 0) {
    int p = n - 1;
    long long mda = abs(a[n - 1].first);
    for (int i = n - 1; i >= 0; i--) {
      if (abs(a[i].first) < mda) {
        mda = a[i].first;
        p = i;
      }
    }
    a[p].first = abs(a[p].first);
    while (k && a[p].first > 0) {
      k--;
      a[p].first -= x;
    }
    if (a[p].first <= 0) a[p].second.second = 1 - a[p].second.second;
    a[p].first = abs(a[p].first);
    ko++;
  }
  for (int i = 0; i < n; i++) a[i].first = abs(a[i].first);
  sort(a.begin(), a.end());
  set<pair<long long, int> > b;
  for (int i = 0; i < n; i++) b.insert(make_pair(a[i].first, i));
  while (k) {
    k--;
    pair<long long, int> temp = *b.begin();
    b.erase(b.begin());
    a[temp.second].first += x;
    b.insert(make_pair(temp.first + x, temp.second));
  }
  vector<long long> ans(n);
  for (int i = 0; i < n; i++)
    ans[a[i].second.first] = a[i].first * (a[i].second.second ? -1 : 1);
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
  puts("");
}
