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.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.PriorityQueue;
import java.util.StringTokenizer;


public class E {

	public static void main(String[] args) throws IOException 
	{
		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);

		int n = sc.nextInt(), k = sc.nextInt(), x = sc.nextInt();
		PriorityQueue<Pair> pq = new PriorityQueue<Pair>();
		int zeros = 0, neg = 0;
		for(int i = 0; i < n; ++i)
		{
			int val = sc.nextInt();
			if(val == 0)
				++zeros;
			else if(val < 0)
				++neg;
			pq.add(new Pair(val, i));
		}
		while(k-->0)
		{
			Pair cur = pq.remove();
			if(zeros > 0)
			{
				--zeros;
				if(neg % 2 == 0)
				{
					++neg;
					cur.val -= x;
				}
				else
					cur.val += x;
			}
			else if(neg % 2 == 0)
			{
				if(cur.val < 0)
					--neg;
				if(cur.val < 0)
					cur.val += x;
				else
					cur.val -= x;
				if(cur.val < 0)
					++neg;
				else if(cur.val == 0)
					++zeros;
			}
			else
			{
				if(cur.val < 0)
					cur.val -= x;
				else
					cur.val += x;
			}
			pq.add(cur);
		}
		long[] b = new long[n];
		for(Pair p: pq)
			b[p.idx] = p.val;
		StringBuilder sb = new StringBuilder();
		for(long bb: b)
			sb.append(bb).append(" ");
		out.println(sb);
		out.flush();
		out.close();
	}
	
	static class Pair implements Comparable<Pair>
	{
		long val;
		int idx;
		
		Pair(long a, int b) { val = a; idx = b; }
		
		public int compareTo(Pair p) 
		{ 
			long diff = Math.abs(val) - Math.abs(p.val);
			if(diff == 0)
				return 0;
			return diff > 0 ? 1 : -1;
		}
	}

	static class Scanner 
	{
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s){	br = new BufferedReader(new InputStreamReader(s));}

		public Scanner(FileReader r){	br = new BufferedReader(r);}

		public String next() throws IOException 
		{
			while (st == null || !st.hasMoreTokens()) 
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public int nextInt() throws IOException {return Integer.parseInt(next());}

		public long nextLong() throws IOException {return Long.parseLong(next());}

		public String nextLine() throws IOException {return br.readLine();}

		public double nextDouble() throws IOException { return Double.parseDouble(next()); }

		public boolean ready() throws IOException {return br.ready();}


	}
}