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;
long long n, k, x;
long long a[211111];
priority_queue<pair<long long, long long> > qm, qp;
queue<long long> qz;
void addM(long long i) { qm.push(make_pair(a[i], i)); }
void addP(long long i) { qp.push(make_pair(-a[i], i)); }
void addZ(long long i) { qz.push(i); }
void add(long long i) {
  if (a[i] < 0)
    addM(i);
  else if (a[i] > 0)
    addP(i);
  else
    addZ(i);
}
void write() {
  for (long long i = 0; i < (long long)(n); i++) {
    cout << a[i] << " ";
  }
  cout << '\n';
}
void addK() {
  while (k) {
    long long m = 1e18;
    long long p = 1e18;
    if (qm.size()) {
      m = -qm.top().first;
    }
    if (qp.size()) {
      p = -qp.top().first;
    }
    long long i;
    if (m < p) {
      i = qm.top().second;
      qm.pop();
      a[i] -= x;
    } else {
      i = qp.top().second;
      qp.pop();
      a[i] += x;
    }
    add(i);
    k--;
  }
}
void inverse() {
  long long m = 1e18;
  long long p = 1e18;
  if (qm.size()) {
    m = -qm.top().first;
  }
  if (qp.size()) {
    p = -qp.top().first;
  }
  long long i;
  long long z;
  if (m < p) {
    i = qm.top().second;
    qm.pop();
    z = +1;
  } else {
    i = qp.top().second;
    qp.pop();
    z = -1;
  }
  while (k) {
    k--;
    a[i] += z * x;
    if (z == 1 && a[i] > 0) break;
    if (z == -1 && a[i] < 0) break;
  }
  add(i);
}
int main(void) {
  cin >> n >> k >> x;
  for (long long i = 0; i < (long long)(n); i++) {
    cin >> a[i];
    add(i);
  }
  if (qz.size() > k) {
    write();
    return 0;
  }
  while (qz.size()) {
    long long i = qz.front();
    qz.pop();
    if (qm.size() % 2 == 0)
      a[i] -= x;
    else
      a[i] += x;
    add(i);
    k--;
  }
  if (qm.size() % 2 == 0) {
    inverse();
  }
  if (qm.size() % 2) {
    addK();
  }
  write();
  return 0;
}
