Problem:
Score : 300 points

Problem Statement

When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l.

You are given N numbers X_1, X_2, ..., X_N, where N is an even number.
For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i.

Find B_i for each i = 1, 2, ..., N.

Constraints

2 \leq N \leq 200000

N is even.

1 \leq X_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
X_1 X_2 ... X_N

Output

Print N lines.
The i-th line should contain B_i.

Sample Input 1

4
2 4 4 3

Sample Output 1

4
3
3
4

Since the median of X_2, X_3, X_4 is 4, B_1 = 4.

Since the median of X_1, X_3, X_4 is 3, B_2 = 3.

Since the median of X_1, X_2, X_4 is 3, B_3 = 3.

Since the median of X_1, X_2, X_3 is 4, B_4 = 4.

Sample Input 2

2
1 2

Sample Output 2

2
1

Sample Input 3

6
5 5 4 4 3 3

Sample Output 3

4
4
4
4
4
4

Solution in C# (.NET Core 3.1.201) (judge verdict: Accepted)

using System;
using System.Numerics;
using System.Linq;
using System.Collections.Generic;


namespace debug
{
    class Program
    {
        static void Main(string[] args)
        {
            ProgramC a = new ProgramC();
            a.main();

        }
    }

    //ABC094
    class ProgramA
    {
        public void main()
        {
            string[] s = Console.ReadLine().Split(' ');
            int a = int.Parse(s[0]);
            int b = int.Parse(s[1]);
            int x = int.Parse(s[2]);

            if(a <= x && x <= a + b)
                Console.WriteLine("YES");
            else
                Console.WriteLine("NO");
        }
    }

    class ProgramB
    {
        public void main()
        {
            string[] s = Console.ReadLine().Split(' ');
            int n = int.Parse(s[0]);
            int m = int.Parse(s[1]);
            int x = int.Parse(s[2]);
            string[] xi = Console.ReadLine().Split(' ');

            int count0 = 0;
            int countn = 0;

            for (int i = 0; i < m; i++)
            {
                if (int.Parse(xi[i]) == x)
                    continue;
                if (int.Parse(xi[i]) < x)
                    count0++;
                else
                    countn++;
            }

            Console.WriteLine(Math.Min(count0,countn));
            

        }
    }

    class ProgramC
    {
        public void main()
        {
            long n = long.Parse(Console.ReadLine());
            string[] xi = Console.ReadLine().Split(' ');

            long[] a = new long[n];

            for (int i = 0; i < n; i++)
                a[i] = long.Parse(xi[i]);

            //ソートする
            Array.Sort(a);

            //
            for (int i = 0; i < n; i++)
            {
                if (a[n / 2 - 1] >= long.Parse(xi[i]))
                    Console.WriteLine(a[n / 2]);
                else
                    Console.WriteLine(a[n / 2 - 1]);
            }
            
            

        }
    }

    class ProgramD
    {
        public void main()
        {
            string[] s = Console.ReadLine().Split(' ');
            long n = long.Parse(s[0]); 
            long k = long.Parse(s[1]);
            string[] t = Console.ReadLine().Split(' ');
            long[] a = new long[n];
            long[] num = new long[40];

            for (int i = 0; i < n; i++)
                a[i] = long.Parse(t[i]);

            //aの和を求める
            for (int i = 0; i < 40; i++)
                for (int j = 0; j < n; j++)
                    num[i] += (a[j] >> i) & 1;

            
            long max = 0;

            //ビット全探索的なやつ
            for(int i = 39;i >= 0;i--)
            {
                long sum = 0;
                //検索の対象の一番頭を考える。これが0ならこれは検索しない
                if ((((k + 1) >> i) & 1) == 0)
                    continue;

                //もし1ならこの桁を0にしてしまえば、その以下は最適な数が出せる
                for (int j = 39; j >= 0; j--)
                {
                    //iを固定したときそれ以上は決まっている
                    if (j > i)
                    {
                        if ((((k + 1) >> j) & 1) == 1)
                            sum += (n - num[j]) * (long)Math.Pow(2, j);
                        else
                            sum += num[j] * (long)Math.Pow(2, j);
                    }
                    else if(i == j)//同じときは0
                    {
                        sum += num[j] * (long)Math.Pow(2, j);
                    }
                    else//ここが最適化できる。1と0の個数が多いほうを残すようにビット選択する
                    {
                        if (n - num[j] > num[j])
                            sum += (n - num[j]) * (long)Math.Pow(2, j);
                        else
                            sum += num[j] * (long)Math.Pow(2, j);
                    }
                }

                //最大更新
                if (sum > max)
                    max = sum;
            }

            //答え出力
            Console.WriteLine(max);
        }

    }

    

}