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.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.PriorityQueue;
import java.util.StringTokenizer;

public class P1 {

	static void print(long[] arr) {
		PrintWriter out = new PrintWriter(System.out);

		out.print(arr[0]);
		for (int i = 1; i < arr.length; i++)
			out.print(" " + arr[i]);
		out.println();
		out.flush();
		out.close();
	}

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

		int n = sc.nextInt();
		long k = sc.nextInt(), x = sc.nextInt();
		long[] arr = new long[n];
		int neg = 0;
		for (int i = 0; i < n; i++) {
			arr[i] = sc.nextInt();
			neg += arr[i] < 0 ? 1 : 0;
		}

		// odd negatives
		if ((neg & 1) == 1) {
			PriorityQueue<Pair> p = new PriorityQueue<Pair>(new Comparator<Pair>() {

				@Override
				public int compare(Pair o1, Pair o2) {
					return (Math.abs(o2.val) - Math.abs(o1.val)) > 0 ? -1
							: (Math.abs(o2.val) - Math.abs(o1.val)) < 0 ? 1 : 0;
				}
			});

			for (int i = 0; i < n; i++)
				p.add(new Pair(arr[i], i));
			while (k-- > 0) {
				Pair cur = p.poll();
				if (cur.val < 0)
					p.add(new Pair(cur.val - x, cur.idx));
				else
					p.add(new Pair(cur.val + x, cur.idx));
			}

			long[] res = new long[n];
			while (!p.isEmpty())
				res[p.peek().idx] = p.poll().val;
			print(res);
			return;
		}

		int best = -1;
		for (int i = 0; i < n; i++) {
			long needed = (Math.abs(arr[i]) + x) / x;
			if (needed <= k && (best == -1 || Math.abs(arr[i]) < Math.abs(arr[best]))) {
				best = i;
			}
		}
		

		// can't make the res negative
		if (best == -1) {
			int minIdx = 0;
			for (int i = 1; i < n; i++) {
				if (Math.abs(arr[i]) < Math.abs(arr[minIdx]))
					minIdx = i;
			}
			if (arr[minIdx] < 0)
				arr[minIdx] += x * k;
			else
				arr[minIdx] -= x * k;
			print(arr);
			return;
		}
		

		// can make a negative from best
		k -= ((Math.abs(arr[best]) + x) / x);
		if (arr[best] < 0)
			arr[best] += x * ((Math.abs(arr[best]) + x) / x);
		else
			arr[best] -= x * ((Math.abs(arr[best]) + x) / x);
		
		PriorityQueue<Pair> p = new PriorityQueue<Pair>(new Comparator<Pair>() {

			@Override
			public int compare(Pair o1, Pair o2) {
				return (Math.abs(o2.val) - Math.abs(o1.val)) > 0 ? -1
						: (Math.abs(o2.val) - Math.abs(o1.val)) < 0 ? 1 : 0;
			}
		});

		for (int i = 0; i < n; i++)
			p.add(new Pair(arr[i], i));
		while (k-- > 0) {
			Pair cur = p.poll();
			if (cur.val < 0)
				p.add(new Pair(cur.val - x, cur.idx));
			else
				p.add(new Pair(cur.val + x, cur.idx));
		}

		long[] res = new long[n];
		while (!p.isEmpty())
			res[p.peek().idx] = p.poll().val;
		print(res);
	}

	static class Pair {

		long val;
		int idx;

		Pair(long v, int i) {
			val = v;
			idx = i;
		}
	}

	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();
		}

		public boolean nextEmpty() throws IOException {
			String s = nextLine();
			st = new StringTokenizer(s);
			return s.isEmpty();
		}
	}
}