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 int MAX_N = (int)2e5 + 123;
const int inf = (int)1e9 + 123;
using namespace std;
set<pair<long long, long long> > first, second;
vector<long long> z;
long long n, k, x;
long long a[MAX_N];
long long ans[MAX_N];
int main() {
  cin >> n >> k >> x;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] > 0)
      first.insert(make_pair(a[i], i));
    else if (a[i] == 0)
      z.push_back(i);
    else
      second.insert(make_pair(-a[i], i));
  }
  while (k > 0 && !z.empty()) {
    k--;
    long long val = z.back();
    z.pop_back();
    if (int((second).size()) % 2 == 0)
      second.insert(make_pair(x, val));
    else
      first.insert(make_pair(x, val));
  }
  if (int((second).size()) % 2 == 0 && k > 0) {
    bool tp = 0;
    if (second.empty() ||
        (!first.empty() && second.begin()->first > first.begin()->first))
      tp = 1;
    if (!tp) {
      pair<long long, long long> now = *second.begin();
      second.erase(second.begin());
      long long mini = min(k * 1ll, now.first / x + 1);
      now.first -= mini * x;
      k -= mini;
      if (now.first < 0)
        first.insert(make_pair(-now.first, now.second));
      else if (now.first == 0)
        z.push_back(now.second);
      else
        second.insert(now);
    } else {
      pair<long long, long long> now = *first.begin();
      first.erase(first.begin());
      long long mini = min(k * 1ll, now.first / x + 1);
      now.first -= mini * x;
      k -= mini;
      if (now.first < 0)
        second.insert(make_pair(-now.first, now.second));
      else if (now.first == 0)
        z.push_back(now.second);
      else
        first.insert(now);
    }
  }
  while (k > 0) {
    k--;
    if (!second.empty() &&
        (first.empty() || second.begin()->first <= first.begin()->first)) {
      pair<long long, long long> now = *second.begin();
      second.erase(second.begin());
      now.first += x;
      second.insert(now);
    } else {
      pair<long long, long long> now = *first.begin();
      first.erase(first.begin());
      now.first += x;
      first.insert(now);
    }
  }
  for (auto i : first) ans[i.second] = i.first;
  for (auto i : second) ans[i.second] = -i.first;
  for (auto i : z) ans[i] = 0;
  for (int i = 1; i <= n; i++) printf("%lld ", ans[i]);
  return 0;
}
