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, m, o, x, y, xx, a[200100], b[200100], negative = 0;
priority_queue<pair<long long, long long>, vector<pair<long long, long long> >,
               greater<pair<long long, long long> > >
    pq;
int main() {
  long long mn = 1e18;
  scanf("%lld %lld %lld", &n, &m, &o);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    if (a[i] < 0) negative++;
  }
  if (negative % 2 == 0) {
    for (int i = 0; i < n; i++) {
      if (abs(a[i]) < mn) {
        mn = abs(a[i]);
        x = i;
      }
    }
    if (a[x] < 0) {
      while (a[x] <= 0) {
        a[x] += o;
        m--;
        if (m == 0) break;
      }
    } else {
      while (a[x] >= 0) {
        a[x] -= o;
        m--;
        if (m == 0) break;
      }
    }
    for (int i = 0; i < n; i++) pq.push(make_pair(abs(a[i]), i));
    for (int i = 0; i < m; i++) {
      x = pq.top().first;
      y = pq.top().second;
      pq.pop();
      if (a[y] < 0)
        xx = x + o;
      else
        xx = x + o;
      pq.push(make_pair(xx, y));
    }
  } else {
    for (int i = 0; i < n; i++) pq.push(make_pair(abs(a[i]), i));
    for (int i = 0; i < m; i++) {
      x = pq.top().first;
      y = pq.top().second;
      pq.pop();
      if (a[y] < 0)
        xx = x + o;
      else
        xx = x + o;
      pq.push(make_pair(xx, y));
    }
  }
  while (!pq.empty()) {
    x = pq.top().first;
    y = pq.top().second;
    pq.pop();
    if (a[y] < 0)
      b[y] = x * -1;
    else
      b[y] = x;
  }
  for (int i = 0; i < n; i++) {
    cout << b[i] << " ";
  }
}
