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>
#pragma warning(disable : 4996)
using namespace std;
int n;
long long x;
long long m;
long long a[300000];
int pm[300000];
int main() {
  int i, j, k, l;
  cin >> n >> m >> x;
  int pro = 1;
  for (i = 0; i < n; i++)
    scanf("%lld", &a[i]), pm[i] = a[i] >= 0 ? 1 : -1, pro *= pm[i];
  if (pro == -1) {
    priority_queue<pair<long long, long long>,
                   vector<pair<long long, long long>>,
                   greater<pair<long long, long long>>>
        pq;
    for (i = 0; i < n; i++) pq.emplace(abs(a[i]), i);
    while (m--) {
      auto tar = pq.top().second;
      pq.pop();
      a[tar] += x * pm[tar];
      pq.emplace(abs(a[tar]), tar);
    }
    for (i = 0; i < n; i++) printf("%lld ", a[i]);
    return 0;
  }
  long long mi = 0x3f3f3f3f3f3f3f3fll, mii;
  for (i = 0; i < n; i++)
    if (mi > abs(a[i])) mi = abs(a[i]), mii = i;
  if (mi - m * x >= 0) {
    for (i = 0; i < n; i++)
      printf("%lld ", i == mii ? a[i] - m * x * pm[i] : a[i]);
    return 0;
  } else {
    a[mii] -= x * (mi / x + 1) * pm[mii];
    pm[mii] *= -1;
    m -= mi / x + 1;
    priority_queue<pair<long long, long long>,
                   vector<pair<long long, long long>>,
                   greater<pair<long long, long long>>>
        pq;
    for (i = 0; i < n; i++) pq.emplace(abs(a[i]), i);
    while (m--) {
      auto tar = pq.top().second;
      pq.pop();
      a[tar] += x * pm[tar];
      pq.emplace(abs(a[tar]), tar);
    }
    for (i = 0; i < n; i++) printf("%lld ", a[i]);
  }
  return 0;
}
