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 javafx.util.Pair;

import java.io.*;
import java.util.*;

/**
 * Created by chao on 16/11/13.
 */
public class A {

    public static void main(String[] args) throws IOException {
        PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));


        String[] nums = br.readLine().split(" ");
        int n = Integer.parseInt(nums[0]);
        int k = Integer.parseInt(nums[1]);
        long x = Long.parseLong(nums[2]);

        boolean positive = true;
        int n0 = 0;
        nums = br.readLine().split(" ");
        PriorityQueue<Node> pq = new PriorityQueue<>();
        for (int i = 0; i < n; i++) {
            long val = Long.parseLong(nums[i]);
            if (val == 0) n0++;
            else if (val < 0) positive = !positive;
            pq.add(new Node(i, val));
        }

        while (k-- > 0) {
            Node nd = pq.poll();
            if (nd.val == 0) {
                if (positive) {
                    nd.val = -x;
                    positive = false;
                } else {
                    nd.val = x;
                }
                n0--;
            } else {
                if (positive) {
                    if (Math.abs(nd.val) == x) {
                        if (nd.val < 0) positive = false;
                        n0++;
                    } else if (Math.abs(nd.val) < x) {
                        positive = false;
                    }
                    nd.val -= getOp(nd.val) * x;

                } else {
                    nd.val += getOp(nd.val) * x;
                }
            }
            pq.add(nd);
        }
        long[] res = new long[n];
        while (!pq.isEmpty()) {
            Node nd = pq.poll();
            res[nd.id] = nd.val;
        }

        for (int i = 0; i < n; i++) {
            if (i != 0) pw.print(' ');
            pw.print(res[i]);
        }
        pw.println();

        pw.flush();
        pw.close();
        br.close();

    }

    private static int getOp(long x){
        if (x > 0) return 1;
        else if (x < 0) return -1;
        else return 0;
    }

}

class Node implements Comparable<Node>{
    int id;
    long val;

    public Node(int id, long val){
        this.id =id;
        this.val = val;
    }

    @Override
    public int compareTo(Node o) {
        long x = Math.abs(val) - Math.abs(o.val);
        if (x > 0) return 1;
        else if (x < 0) return -1;
        else return 0;
    }
}

