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.*;
import java.io.*;

public class D {
	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

		String[] line = br.readLine().split(" ");
		int N = Integer.parseInt(line[0]);
		int K = Integer.parseInt(line[1]);
		int X = Integer.parseInt(line[2]);

		line = br.readLine().split(" ");

		PriorityQueue<Num> pq = new PriorityQueue<>();

		int nNeg = 0;
		long[] vals = new long[N];
		for (int i = 0; i < N; i++) {
			vals[i] = Integer.parseInt(line[i]);
			if (vals[i] < 0) {
				nNeg++;
			}
			pq.add(new Num(vals[i], i));
		}

		for (int i = 0; i < K; i++) {
			Num n = pq.poll();
			long nn = n.val;
			if (nNeg % 2 == 1) {
				if (nn >= 0) {
					nn += X;	
				}	else {
					nn -= X;
				}
			}	else {
				if (nn >= 0) {
					nn -= X;	
					if (nn < 0) nNeg++;
				}	else {
					nn += X;
					if (nn >= 0) nNeg--;
				}
			}

//			System.err.println("Set " + n.idx + " to " + nn);	
			vals[n.idx] = nn;			
			pq.add(new Num(nn, n.idx));
		}

		StringBuilder out = new StringBuilder();

		int added = 0;
		for (long i : vals) {
			out.append(i); 
			added++;
			if (added < N) {
				out.append(" ");
			}
		}

		System.out.println(out);
		
	}
}

class Num implements Comparable<Num> {
	long val;
	int idx;
	
	Num(long val, int idx) {
		this.val = val;
		this.idx = idx;
	}	

	public int compareTo(Num other) {
		if (Math.abs(this.val) > Math.abs(other.val)) {
			return 1;
		}

		if (Math.abs(this.val) == Math.abs(other.val)) return 0;
		return -1;
	}
}
