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() {
  long long int n, k, x;
  cin >> n >> k >> x;
  vector<long long int> v1(n);
  for (auto &it : v1) cin >> it;
  long long int neg = 0;
  long long int pos = 0;
  for (long long int i = 0; i <= n - 1; i++) {
    if (v1[i] < 0)
      neg++;
    else
      pos++;
  }
  vector<long long int> v2(n);
  if (neg % 2 == 0) {
    for (long long int i = 0; i <= n - 1; i++) v2[i] = abs(v1[i]);
    long long int a1 = distance(v2.begin(), min_element(v2.begin(), v2.end()));
    if (v1[a1] >= 0) {
      while (1) {
        if (v1[a1] < 0 || k == 0) break;
        v1[a1] -= x;
        k--;
      }
    } else if (v1[a1] < 0) {
      while (1) {
        if (v1[a1] > 0 || k == 0) break;
        v1[a1] += x;
        k--;
      }
    }
  }
  for (long long int i = 0; i <= n - 1; i++) v2[i] = abs(v1[i]);
  multimap<long long int, long long int> m1;
  for (long long int i = 0; i <= n - 1; i++) m1.insert(make_pair(v2[i], i));
  for (long long int i = 0; i <= k - 1; i++) {
    multimap<long long int, long long int>::iterator itr;
    itr = m1.begin();
    long long int a1 = itr->first;
    long long int a2 = itr->second;
    m1.erase(itr);
    m1.insert(make_pair(a1 + x, a2));
  }
  multimap<long long int, long long int> m2;
  vector<long long int> ans;
  for (auto &it : m1) m2.insert(make_pair(it.second, it.first));
  for (auto &it : m2) ans.push_back(it.second);
  for (long long int i = 0; i <= n - 1; i++)
    if (v1[i] < 0) ans[i] *= -1;
  for (long long int i = 0; i <= n - 1; i++) cout << ans[i] << " ";
}
