Problem: Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.

Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.

There are two conditions though: 

  * For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x. 
  * The modified rating changes must be perfectly balanced — their sum must be equal to 0. 



Can you help with that?

Input

The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.

Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.

The sum of all a_i is equal to 0.

Output

Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.

For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.

If there are multiple solutions, print any. We can show that a solution exists for any valid input.

Examples

Input


3
10
-5
-5


Output


5
-2
-3


Input


7
-7
-29
0
3
24
-29
38


Output


-3
-15
0
2
12
-15
19

Note

In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.

In the second example there are 6 possible solutions, one of them is shown in the example output.

Solution (in JAVA):

Solution:
import java.util.Scanner;
import java.util.Arrays;
import java.util.ArrayList;
import java.util.*;
import java.util.List;
import java.lang.Math;
import java.util.HashMap;
public class javasample {
	public static int useBs(char[] arr, char targetValue) {	
		int a =  Arrays.binarySearch(arr, targetValue);
		if(a >=0 && a<arr.length)
			return a;
		else
			return -1;
	}
	public static void printArray(int []a) {
		for(int i=0;i<a.length;i++) {
			System.out.println(a[i]+" ");
		}
	}
	static boolean isPalindrome(String s) {
		int l=s.length();
		for(int i=0;i<l/2;i++)
		{
			if(s.charAt(i)!=s.charAt(l-i-1))
				return false;
		}
		return true;
	}
	static String revString(String str)
	{
		return new StringBuffer(str).reverse().toString();
	}
	static String sortString(String s) {
		char a[]=s.toCharArray();
		Arrays.sort(a);
		return new String(a);
	}
	static int sumList(List<Integer>a) {
		return a.stream().mapToInt(Integer::intValue).sum();
	}
    public static void main(String[]args) {
    	Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		int a[]=new int[t];
		List<Integer>lst1=new ArrayList<Integer>();
		List<Integer>lst2=new ArrayList<Integer>();
		for(int i=0;i<t;i++) {
			a[i]=sc.nextInt();
			if(a[i]%2==0) {
				lst1.add(a[i]/2);
				lst2.add(a[i]/2);
			}else {
				if(a[i]<0) {
					lst1.add(((int)a[i]/2)-1);
					lst2.add((int)a[i]/2);
				}else {
					lst1.add((int)a[i]/2);
					lst2.add(((int)a[i]/2)+1);
				}
			}
		}
		int x=sumList(lst1);
		int y=sumList(lst2);
		int i=0,cnt=0;
		while(true) {
			if(lst1.get(i)<lst2.get(i)) {
			lst1.set(i, lst2.get(i));
			cnt+=1;}
			i+=1;
			if(cnt==Math.abs(x))break;
		}
		for( i=0;i<t;i++) {
			System.out.println(lst1.get(i));
		}
    }
}