Nick had received an awesome array of integers a=[a_1, a_2, ..., a_n] as a gift for his 5 birthday from his mother. He was already going to explore its various properties but after unpacking he was disappointed a lot because the product a_1 ⋅ a_2 ⋅ ... a_n of its elements seemed to him not large enough.

He was ready to throw out the array, but his mother reassured him. She told him, that array would not be spoiled after the following operation: choose any index i (1 ≤ i ≤ n) and do a_i := -a_i - 1.

For example, he can change array [3, -1, -4, 1] to an array [-4, -1, 3, 1] after applying this operation to elements with indices i=1 and i=3. 

Kolya had immediately understood that sometimes it's possible to increase the product of integers of the array a lot. Now he has decided that he wants to get an array with the maximal possible product of integers using only this operation with its elements (possibly zero, one or more times, as many as he wants), it is not forbidden to do this operation several times for the same index. 

Help Kolya and print the array with the maximal possible product of elements a_1 ⋅ a_2 ⋅ ... a_n which can be received using only this operation in some order.

If there are multiple answers, print any of them.

Input

The first line contains integer n (1 ≤ n ≤ 10^{5}) — number of integers in the array.

The second line contains n integers a_1, a_2, …, a_n (-10^{6} ≤ a_i ≤ 10^{6}) — elements of the array

Output

Print n numbers — elements of the array with the maximal possible product of elements which can be received using only this operation in some order from the given array.

If there are multiple answers, print any of them.

Examples

Input


4
2 2 2 2


Output


-3 -3 -3 -3 

Input


1
0


Output


0 

Input


3
-3 -3 2


Output


-3 -3 2 
import java.util.*;
import java.io.*;
import java.math.*;

public class round569d2b {

	public static void main(String args[]) {
		FastScanner in = new FastScanner(System.in);
		
		int n = in.nextInt();
		int[] arr = new int[n];
		for (int i = 0; i < n; i++) {
			arr[i] = in.nextInt();
		}
		if (n % 2 == 0) {
			for (int i = 0; i < n; i++) {
				if (arr[i] >= 0) {
					arr[i] = -1*arr[i]-1;
				}
			}
		}
		else {
			int max = Integer.MIN_VALUE;
			int maxIndex = 0;
			for (int i = 0; i < n; i++) {
				int elem = arr[i];
				if (elem < 0) {
					elem = -1*elem-1;
				}
				if (elem > max) {
					max = elem;
					maxIndex = i;
				}
			}
			for (int i = 0; i < n; i++) {
				if (i == maxIndex) {
					if (arr[i] < 0) {
						arr[i] = -1*arr[i]-1;
					}
				}
				else {
					if (arr[i] >= 0) {
						arr[i] = -1*arr[i]-1;
					}
				}
			}
		}
		StringBuilder sb = new StringBuilder();
		for (int i = 0; i < n ;i++) {
			sb.append(arr[i] + " ");
			
		}
		System.out.println(sb);

	}

	// ======================================================================================
	// =============================== Reference Code =======================================
	// ======================================================================================

	static int greatestDivisor(int n) {
		int limit = (int) Math.sqrt(n);
		int max = 1;
		for (int i = 2; i <= limit; i++) {
			if (n % i == 0) {
				max = Integer.max(max, i);
				max = Integer.max(max, n / i);
			}
		}
		return max;
	}

	// Method to return all primes smaller than or equal to 
	// n using Sieve of Eratosthenes 
	static boolean[] sieveOfEratosthenes(int n) {
		// Create a boolean array "prime[0..n]" and initialize 
		// all entries it as true. A value in prime[i] will 
		// finally be false if i is Not a prime, else true. 
		boolean prime[] = new boolean[n + 1];
		for (int i = 0; i <= n; i++)
			prime[i] = true;

		prime[0] = false;
		prime[1] = false;

		for (int p = 2; p * p <= n; p++) {
			// If prime[p] is not changed, then it is a prime 
			if (prime[p] == true) {
				// Update all multiples of p 
				for (int i = p * p; i <= n; i += p)
					prime[i] = false;
			}
		}

		return prime;
	}

	// Binary search for number greater than or equal to target
	// returns -1 if number not found
	private static int bin_gteq(int[] a, int key) {
		int low = 0;
		int high = a.length;
		int max_limit = high;
		while (low < high) {
			int mid = low + (high - low) / 2;
			if (a[mid] < key) {
				low = mid + 1;
			} else
				high = mid;
		}

		return high == max_limit ? -1 : high;
	}

	public static int gcd(int a, int b) {
		if (a == 0)
			return b;

		return gcd(b % a, a);
	}

	static class Tuple<X, Y> {
		public final X x;
		public final Y y;

		public Tuple(X x, Y y) {
			this.x = x;
			this.y = y;
		}

		public String toString() {
			return "(" + x + "," + y + ")";
		}
	}

	static class Tuple3<X, Y, Z> {
		public final X x;
		public final Y y;
		public final Z z;

		public Tuple3(X x, Y y, Z z) {
			this.x = x;
			this.y = y;
			this.z = z;
		}

		public String toString() {
			return "(" + x + "," + y + "," + z + ")";
		}
	}

	static Tuple3<Integer, Integer, Integer> gcdExtended(int a, int b, int x, int y) {
		// Base Case 
		if (a == 0) {
			x = 0;
			y = 1;
			return new Tuple3(0, 1, b);
		}

		int x1 = 1, y1 = 1; // To store results of recursive call 
		Tuple3<Integer, Integer, Integer> tuple = gcdExtended(b % a, a, x1, y1);
		int gcd = tuple.z;
		x1 = tuple.x;
		y1 = tuple.y;

		// Update x and y using results of recursive 
		// call 
		x = y1 - (b / a) * x1;
		y = x1;

		return new Tuple3(x, y, gcd);
	}

	// Returns modulo inverse of a  
	// with respect to m using extended 
	// Euclid Algorithm. Refer below post for details: 
	// https://www.geeksforgeeks.org/multiplicative-inverse-under-modulo-m/ 
	static int inv(int a, int m) {
		int m0 = m, t, q;
		int x0 = 0, x1 = 1;

		if (m == 1)
			return 0;

		// Apply extended Euclid Algorithm 
		while (a > 1) {
			// q is quotient 
			q = a / m;

			t = m;

			// m is remainder now, process 
			// same as euclid's algo 
			m = a % m;
			a = t;

			t = x0;

			x0 = x1 - q * x0;

			x1 = t;
		}

		// Make x1 positive 
		if (x1 < 0)
			x1 += m0;

		return x1;
	}

	// k is size of num[] and rem[]. 
	// Returns the smallest number 
	// x such that: 
	// x % num[0] = rem[0], 
	// x % num[1] = rem[1], 
	// .................. 
	// x % num[k-2] = rem[k-1] 
	// Assumption: Numbers in num[] are pairwise  
	// coprime (gcd for every pair is 1) 
	static int findMinX(int num[], int rem[], int k) {
		// Compute product of all numbers 
		int prod = 1;
		for (int i = 0; i < k; i++)
			prod *= num[i];

		// Initialize result 
		int result = 0;

		// Apply above formula 
		for (int i = 0; i < k; i++) {
			int pp = prod / num[i];
			result += rem[i] * inv(pp, num[i]) * pp;
		}

		return result % prod;
	}

	/**
	 * Source: Matt Fontaine
	 */
	static class FastScanner {
		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar;
		private int chars;

		public FastScanner(InputStream stream) {
			this.stream = stream;
		}

		int read() {
			if (chars == -1)
				throw new InputMismatchException();
			if (curChar >= chars) {
				curChar = 0;
				try {
					chars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (chars <= 0)
					return -1;
			}
			return buf[curChar++];
		}

		boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		boolean isEndline(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}

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

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

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

		public String next() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}

		public String nextLine() {
			int c = read();
			while (isEndline(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isEndline(c));
			return res.toString();
		}
	}

}