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 mod = 998244353;
const int maxn = 200000 + 5;
int n, k, x;
long long a[maxn];
int main() {
  scanf("%d%d%d", &n, &k, &x);
  int zs = 0, ze = 0, fs = 0, tot = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", a + i);
    if (a[i] == 0)
      ze++;
    else if (a[i] > 0)
      zs++;
    else
      fs++;
  }
  if (ze) {
    if (ze > k) {
      printf("%I64d ", a[1] + 1ll * k * x);
      for (int i = 2; i <= n; i++) printf("%I64d ", a[i]);
      return 0;
    }
    int last = 0;
    for (int i = 1; i <= n; i++) {
      if (a[i]) continue;
      a[i] += x;
      last = i;
    }
    if (fs % 2 == 0) a[last] = -x;
    tot = ze;
  }
  zs = 0, fs = 0;
  for (int i = 1; i <= n; i++)
    if (a[i] > 0)
      zs++;
    else
      fs++;
  if (fs % 2 == 0) {
    int m = 1e9 + 7, p = -1;
    for (int i = 1; i <= n; i++) {
      if (abs(a[i]) < m) m = abs(a[i]), p = i;
    }
    int mn = a[p];
    int cs = abs(m) / x + 1;
    if (cs + tot > k) {
      int rest = k - tot;
      if (a[p] < 0)
        a[p] += 1ll * rest * x;
      else
        a[p] -= 1ll * rest * x;
      tot = k;
    } else {
      if (a[p] < 0)
        a[p] += 1ll * cs * x;
      else
        a[p] -= 1ll * cs * x;
      tot += cs;
    }
  }
  set<pair<long long, int> > st;
  for (int i = 1; i <= n; i++) st.insert({abs(a[i]), i});
  for (; tot < k; tot++) {
    auto it = st.begin();
    int id = it->second;
    if (a[id] < 0)
      a[id] -= x;
    else
      a[id] += x;
    st.erase(it);
    st.insert({abs(a[id]), id});
  }
  for (int i = 1; i <= n; i++) printf("%I64d ", a[i]);
  return 0;
}
