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.*;
import java.util.*;

public class D {
	public static void main(String[] args) throws Exception {
		BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);
		Scanner s = new Scanner(System.in);
		
		String[] split = f.readLine().split("\\s+");
		int n = Integer.parseInt(split[0]), k = Integer.parseInt(split[1]); long x = Integer.parseInt(split[2]);
		long[] a = new long[n];
		TreeSet<Num> positive = new TreeSet<>(), negative = new TreeSet<>(), zero = new TreeSet<>();
		split = f.readLine().split("\\s+");
		for(int i = 0; i < n; i++) {
			a[i] = Integer.parseInt(split[i]);
			if(a[i] < 0) {
				negative.add(new Num(i, a[i]));
			} else if(a[i] > 0) {
				positive.add(new Num(i, a[i]));
			} else {
				zero.add(new Num(i, 0));
			}
		}
		//If number of negatives is even, we wanna drag the smallest positive number into the negatives.
		//If there are zeros and an even number of negatives, we wanna bring one zero into the negatives, the rest positive.
		//If zeros and no odd negatives, make all zeros positive.
		//Once there are no zeros and there is an odd number of negatives, increase the number that has the smallest absolute value until the end.
		for(int i = 0; i < k; i++) {
			if(negative.size()%2==1) {
				if(!zero.isEmpty()) {
					Num now = zero.pollFirst();
					now.val += x;
					if(now.val < 0) negative.add(now);
					else if(now.val > 0) positive.add(now);
					else zero.add(now);
				} else {
					if(!positive.isEmpty() && (negative.isEmpty() || positive.first().val <= Math.abs(negative.first().val))) {
						Num now = positive.pollFirst();
						now.val += x;
						if(now.val < 0) negative.add(now);
						else if(now.val > 0) positive.add(now);
						else zero.add(now);
					} else if(!negative.isEmpty()) {
						Num now = negative.pollFirst();
						now.val -= x;
						if(now.val < 0) negative.add(now);
						else if(now.val > 0) positive.add(now);
						else zero.add(now);
					}
				}
			} else {
				if(!zero.isEmpty()) {
					Num now = zero.pollFirst();
					now.val -= x;
					if(now.val < 0) negative.add(now);
					else if(now.val > 0) positive.add(now);
					else zero.add(now);
				} 
				else if(!positive.isEmpty() && (negative.isEmpty() || positive.first().val <= Math.abs(negative.first().val))) {
					Num now = positive.pollFirst();
					now.val -= x;
					if(now.val < 0) negative.add(now);
					else if(now.val > 0) positive.add(now);
					else zero.add(now);
				} else if(!negative.isEmpty()) {
					Num now = negative.pollFirst();
					now.val += x;
					if(now.val < 0) negative.add(now);
					else if(now.val > 0) positive.add(now);
					else zero.add(now);
				}
			}
		}
		Num now = null;
		while(!positive.isEmpty()) a[(now = positive.pollFirst()).index] = now.val;
		while(!negative.isEmpty()) a[(now = negative.pollFirst()).index] = now.val;
		while(!zero.isEmpty()) a[(now = zero.pollFirst()).index] = now.val;
		
		for(int i = 0; i < n; i++) out.print(a[i] + " ");
		out.println();
		out.close();
	}
}

class Num implements Comparable<Num>{
	int index; long val;
	
	public Num(int i, long v) {
		index=  i; val = v;
	}

	public int compareTo(Num o) {
		long abs1 = Math.abs(val), abs2 = Math.abs(o.val);
		if(abs1 == abs2) return index - o.index;
		return abs1 < abs2 ? -1 : 1;
	}
	
	public String toString() {
		return val + " " + index;
	}
}