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 long long maxn = 2e5 + 5;
long long a[maxn];
bool ne[maxn];
int main() {
  long long n, k, x;
  scanf("%lld %lld %lld", &n, &k, &x);
  long long mi = INT_MAX;
  long long pos = -1;
  bool f = 0;
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    if (a[i] < 0) {
      a[i] = -a[i];
      ne[i] = 1;
      f ^= 1;
    }
    if (a[i] < mi) {
      mi = a[i];
      pos = i;
    }
  }
  if (f == 1) {
    set<pair<long long, long long> > st;
    for (long long i = 1; i <= n; i++) st.insert(make_pair(a[i], i));
    while (k--) {
      long long p = st.begin()->second;
      st.erase(st.begin());
      a[p] += x;
      st.insert(make_pair(a[p], p));
    }
    for (long long i = 1; i <= n; i++) {
      if (ne[i])
        printf("%lld ", -a[i]);
      else
        printf("%lld ", a[i]);
    }
    return 0;
  }
  if (mi <= x * k) {
    long long t = mi / x;
    mi -= t * x;
    k -= t;
    if (k) {
      k--;
      mi -= x;
    }
    a[pos] = -mi;
    ne[pos] ^= 1;
    set<pair<long long, long long> > st;
    for (long long i = 1; i <= n; i++) st.insert(make_pair(a[i], i));
    while (k--) {
      long long p = st.begin()->second;
      st.erase(st.begin());
      a[p] += x;
      st.insert(make_pair(a[p], p));
    }
    for (long long i = 1; i <= n; i++) {
      if (ne[i])
        printf("%lld ", -a[i]);
      else
        printf("%lld ", a[i]);
    }
  } else {
    a[pos] -= x * k;
    for (long long i = 1; i <= n; i++)
      if (ne[i])
        printf("%lld ", -a[i]);
      else
        printf("%lld ", a[i]);
  }
  return 0;
}
