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.PriorityQueue;
import java.util.Scanner;

public class ProblemD {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner input = new Scanner(System.in);
		int n = input.nextInt();
		int k = input.nextInt();
		long x = input.nextInt();
		long[] arr = new long[n];
		boolean positive = true;
		for(int a = 0; a < n; a++){
			arr[a] = input.nextInt();
			if(arr[a] < 0){
				positive = !positive;
			}
		}
		PriorityQueue<Point> points = new PriorityQueue<Point>();
		for(int a = 0; a < n;a ++){
			Point p = new Point(arr[a], a);
			points.add(p);
		}
		int count = 0;
		int mult = positive?-1:1;
		while(true){
			if(count == k){
				break;
			}
			count++;
			Point p = points.poll();
			long prev = p.val;
			p.val = p.val + mult*((Math.abs(p.val) == 0)?((p.negativeStart)?-1:1):p.val/Math.abs(p.val))*x;
			arr[p.id] = p.val;
			if((prev <= 0 && p.val > 0) ||(prev >= 0 && p.val < 0)){
				mult = 1;
			}
			points.add(p);
		}
		for(int a = 0; a < n; a++){
			System.out.print(arr[a] + " ");
		}
	}

	public static class Point implements Comparable<Point> {
		long val;
		int id;
		boolean negativeStart;

		public Point(long _val, int _id) {
			val = _val;
			id = _id;
			negativeStart = val < 0;
		}

		public int compareTo(Point o) {
			return Long.compare(Math.abs(val), Math.abs(o.val));
		}
	}
}
