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 
//package Chapter2;

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

public class MaximAndArray2 {
	static class pair implements Comparable<pair>{
		int pos;
		long val;
		public pair(int p, long v) {
			val = v;
			pos = p;
		}
		@Override
		public int compareTo(pair p) {
			return val-p.val>0?1:val==p.val?pos-p.pos:-1;
		}
		
	}
	public static void main(String[] args) throws IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(in.readLine());
		PrintWriter out = new PrintWriter(System.out);
		int n = Integer.parseInt(st.nextToken()), k = Integer.parseInt(st.nextToken());
		long x = Integer.parseInt(st.nextToken());
		long elm[] = new long[n];
		int countZeros = 0, countNeg = 0;
		long min = (long) 1e10;
		int pos = -1;
		st = new StringTokenizer(in.readLine());
		for (int i = 0; i < elm.length; i++) {
			elm[i] = Long.parseLong(st.nextToken());
			if (elm[i] == 0)
				countZeros++;
			else if (elm[i] < 0)
				countNeg++;
			if (Math.abs(elm[i]) < min) {
				min = Math.abs(elm[i]);
				pos = i;
			}
		}
		boolean foundZeros = countZeros > 0 ? true : false;
		if (countZeros > k) {
			for (int i = 0; i < k; i++) {
				elm[i] += x;
			}
			k = 0;
		} else if (countZeros == k) {

			// need one negative zero;
			for (int i = 0; i < elm.length; i++) {
				if (elm[i] == 0) {
					if (countZeros == 1) {
						if (countNeg % 2 == 0) {
							elm[i] -= x;
						} else
							elm[i] += x;
					} else
						elm[i] += x;

					countZeros--;
				}
			}
			k = 0;
		}else if(countZeros!=0) {
			k-=countZeros;
			if(countNeg%2==0) {
				for (int i = 0; i < elm.length; i++) {
					if(elm[i]==0) {
						if(countZeros==1)
							elm[i]-=x;
						else
							elm[i]+=x;
						countZeros--;
					}
				}
			}else {
				for (int i = 0; i < elm.length; i++) {
					if(elm[i]==0) {
						elm[i]+=x;
					}
				}
			}
		}
		if(!foundZeros) {
			if(countNeg%2==0) {
				//need to change the sign of one of them
				if(elm[pos]<0) {
					while(elm[pos]<=0&&k>0) {
						elm[pos]+=x;
						k--;
					}
				}else {
					while(elm[pos]>=0&&k>0) {
						elm[pos]-=x;
						k--;
					}
				}
			}
		}
		
		PriorityQueue<pair> pq = new PriorityQueue<pair>();
		for (int i = 0; i < elm.length; i++) {
			pq.add(new pair(i,Math.abs(elm[i])));
		}
		while(k-->0) {
			pair cur = pq.remove();
			cur.val+=x;
			pq.add(cur);
		}
		while(!pq.isEmpty()) {
			pair cur = pq.remove();
			if(elm[cur.pos]>0)
				elm[cur.pos]=cur.val;
			else
				elm[cur.pos]=cur.val*-1;
		}
		
		for (int i = 0; i < elm.length; i++) {
			out.print(elm[i]+" ");
		}
		out.flush();out.close();
	}
}
