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.util.Arrays;
import java.util.Scanner;
import java.util.TreeSet;
import java.util.stream.Collectors;
import java.util.stream.Stream;

public class ComplexTask {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int k = scanner.nextInt();
        long x = scanner.nextLong();

        TreeSet<ItemAndPos> set = new TreeSet<>( (a, b) -> {
            int f = Long.compare(Math.abs(a.value), Math.abs(b.value));
            if (f == 0) {
                return a.pos - b.pos;
            }
            return f;
        });
        int negarives = 0;
        for (int i = 0; i < n; i++) {
            long val = scanner.nextLong();
            set.add(new ItemAndPos(val, i));
            if (val < 0) {
                negarives++;
            }
        }

        if (negarives % 2 == 0) {
            ItemAndPos i = set.pollFirst();
            int kk = (int) (Math.abs(i.value) / x) + 1;
            int mk = Math.min(kk, k);
            if (i.value < 0) {
                i.value += mk * x;
            } else {
                i.value -= mk * x;
            }
            k -= mk;
            set.add(i);
        }

        int current = 0;
        while (k > 0) {
            ItemAndPos i = set.pollFirst();
            if (i.value < 0) {
                i.value -= x;
            } else {
                i.value += x;
            }
            set.add(i);
            k--;
        }
        String[] arr2 = new String[n];
        for (ItemAndPos i : set) {
            arr2[i.pos] = "" + i.value;
        }
        System.out.println(Stream.of(arr2).collect(Collectors.joining(" ")));
    }


    static class ItemAndPos {
        long value;
        int pos;

        public ItemAndPos(long value, int pos) {
            this.value = value;
            this.pos = pos;
        }
    }

}
