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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <algorithm>
#include <cmath>
#include <deque>
#include <iostream>
#include <numeric>
#include <sstream>
#include <string>
#include <string.h>
#include <vector>

#define REP(i,x) for(int i{ 0 }; i < (int)(x); i++)
#define REPC(i,x) for(int i{ 0 }; i <= (int)(x); i++)
#define RREP(i,x) for(int i{ (int)(x) - 1 }; i >= 0 ;i--)
#define RREPC(i,x) for(int i{ (int)(x)}; i >= 0; i--)
#define REP1O(i,x) for(int i{ 1 }; i < (int)(x); i++)
#define REP1C(i,x) for(int i{ 1 }; i <= (int)(x); i++)

#define PB push_back
#define MP make_pair
#define F first
#define S second
#define SZ(x) ((int)(x).size())
#define ALL(x) (x).begin(),(x).end()

typedef int64_t ll;

const int dx[4] = { 1, 0, -1,  0 };
const int dy[4] = { 0, 1,  0, -1 };
const int INTMAX = 2147483647;
const ll LLMAX = 9223372036854775807;

template<class T>bool chmax(T& a, const T& b) { if (a < b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T& a, const T& b) { if (b < a) { a = b; return 1; } return 0; }
void SWAP(ll& a, ll& b) { a ^= b; b ^= a; a ^= b; }
void SWAP(int& a, int& b) { a ^= b; b ^= a; a ^= b; }

using namespace std;


class C {
	int n;
	vector<int> x;
public:
	C()
	{
		cin >> n;
		x.resize(n);
		REP(i, n)
			cin >> x[i];
	}

	void solve()
	{

		REP(i, n / 2)
			cout << x[i / 2] << endl;
		REP(i, n / 2)
			cout << x[i / 2 - 1] << endl;

	}
};


int main()
{
	C solution;
	solution.solve();

	return 0;
}