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

public class MaximAndArray_CF721D {

	
	
	
	public static void main(String[] args) throws IOException {

		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		
		int n = sc.nextInt();
		int k = sc.nextInt();
		int x = sc.nextInt();
		
		boolean neg = false;
		PriorityQueue<Num> pq = new PriorityQueue<Num>();
		Num [] arr = new Num [n];
		for(int i = 0; i < n; i++) {
			int y = sc.nextInt();
			if( y < 0)
				neg = !neg;
			Num num = new Num(i,y);
			arr[i] = num;
			pq.add(num);
		}
		
		if(!neg) {
			Num num = pq.remove();
			if(num.num < 0)
				while(num.num <= 0 && k > 0) {
					num.num += x;
					k--;
				}
			else
				while(num.num >= 0 && k > 0) {
					num.num -= x;
					k--;
				}
			pq.add(num);
		}
		
		while(k > 0) {
			k--;
			Num num = pq.remove();
			num.add(x);
			pq.add(num);
		}
		out.print(arr[0].num);
		for(int i = 1; i < n; i++)
			out.print(" " + arr[i].num);
		
		out.flush();
		out.close();
	}
	
	static class Num implements Comparable<Num>{
		int ind;long num;
		public Num(int ind, int num) {
			this.ind = ind;
			this.num = num;
		}
		
		public int compareTo(Num o) {
			return Math.abs(num) == Math.abs(o.num)? ind - o.ind: Math.abs(num) < Math.abs(o.num)? -1:1;
		}
		
		public void add(int x) {
			if(num < 0)
				num -= x;
			else
				num += x;
		}
	}
	
	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

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

		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 {
			String x = next();
			StringBuilder sb = new StringBuilder("0");
			double res = 0, f = 1;
			boolean dec = false, neg = false;
			int start = 0;
			if (x.charAt(0) == '-') {
				neg = true;
				start++;
			}
			for (int i = start; i < x.length(); i++)
				if (x.charAt(i) == '.') {
					res = Long.parseLong(sb.toString());
					sb = new StringBuilder("0");
					dec = true;
				} else {
					sb.append(x.charAt(i));
					if (dec)
						f *= 10;
				}
			res += Long.parseLong(sb.toString()) / f;
			return res * (neg ? -1 : 1);
		}

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

	}

}