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 OO = (int)1e9;
const double eps = 1e-10;
const int MX = 2e5 + 5;
int n, x, k, negCnt, mn = 1e9 + 5, mnIdx;
long long a[MX];
bool neg[MX];
set<pair<long long, int>> S;
void solve() {
  for (int i = 0; i < int(n); ++i) S.insert({a[i], i});
  while (k--) {
    long long mn = S.begin()->first;
    int idx = S.begin()->second;
    S.erase(S.begin());
    a[idx] += x;
    S.insert({mn + x, idx});
  }
  for (int i = 0; i < int(n); ++i) cout << a[i] * (neg[i] ? -1 : 1) << ' ';
}
void print() {
  cout << "> "
       << "k"
       << " = " << (k) << '\n';
  for (int i = 0; i < int(n); ++i) cout << a[i] << '\t';
  cout << '\n';
  for (int i = 0; i < int(n); ++i) cout << neg[i] << '\t';
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cout.precision(10);
  cin >> n >> k >> x;
  for (int i = 0; i < int(n); ++i) {
    cin >> a[i];
    if (a[i] < 0) neg[i] = true, ++negCnt;
    a[i] = abs(a[i]);
    if (a[i] < mn) mn = a[i], mnIdx = i;
  }
  if (negCnt & 1)
    solve();
  else {
    int steps = min((mn + x - 1) / x, k);
    a[mnIdx] -= steps * x;
    k -= steps;
    if (!a[mnIdx] && k) --k, a[mnIdx] -= x;
    if (a[mnIdx] < 0) neg[mnIdx] = !neg[mnIdx];
    a[mnIdx] = abs(a[mnIdx]);
    solve();
  }
  return 0;
}
