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 
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

/**
 * Created by pallavi on 17/9/16.
 */
public class D721 {
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer tokenizer = new StringTokenizer(reader.readLine());
        int n = Integer.parseInt(tokenizer.nextToken());
        int k = Integer.parseInt(tokenizer.nextToken());
        long x = Integer.parseInt(tokenizer.nextToken());

        PriorityQueue<Pair> queue = new PriorityQueue<>();
        tokenizer = new StringTokenizer(reader.readLine());
        boolean ng = false;
        for (int i = 0; i < n; i++) {
            long l = Integer.parseInt(tokenizer.nextToken());
            queue.add(new Pair(l, i));
            if (l < 0) {
                ng ^= true;
            }
        }
        for (int i = 0; i < k; i++) {
            Pair pair = queue.poll();
            long l = pair.n;
            if (ng) {
                if (l < 0) {
                    l -= x;
                } else {
                    l += x;
                }
            } else {
                if (l < 0) {
                    l += x;
                    if (l >= 0) {
                        ng = true;
                    }
                } else {
                    l -= x;
                    if (l < 0) {
                        ng = true;
                    }
                }
            }
            queue.add(new Pair(l, pair.i));
        }
        long[] ord = new long[n];
        for (Pair p: queue) {
            ord[p.i] = p.n;
        }
        for (int i = 0; i < n; i++) {
            System.out.print(ord[i] + " ");
        }
    }
    private static class Pair implements Comparable<Pair> {
        long n;
        int i;

        public Pair(long n, int i) {
            this.n = n;
            this.i = i;
        }

        @Override
        public int compareTo(Pair pair) {
            if (Math.abs(this.n) < Math.abs(pair.n)) return -1;
            else if (Math.abs(this.n) > Math.abs(pair.n)) return 1;
            else return this.i - pair.i;
        }
    }
}
