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: Accepted)

#ifndef BZ
#pragma GCC optimize "-O3"
#endif
#include <bits/stdc++.h>

#define FASTIO

#ifdef FASTIO
#define scanf abacaba
#define printf abacaba
#endif

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

using namespace std;


/*
ll pw(ll a, ll b) {
	ll ans = 1; while (b) {
		while (!(b & 1)) b >>= 1, a = (a * a) % MOD;
		ans = (ans * a) % MOD, --b;
	} return ans;
}
*/

#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
typedef tree<pair<int, int>,
		null_type,
		less<pair<int, int>>,
		rb_tree_tag,
		tree_order_statistics_node_update>
		st;

//find_by_order(k);
//order_of_key(x);


const int MAXN = 220000;
int a[MAXN];

int main() {
#ifdef FASTIO
	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
#endif
	st ss;
	int n;
	cin >> n;
	for (int i = 0; i < n; ++i) {
		cin >> a[i];
		ss.insert(make_pair(a[i], i));
	}
	for (int i = 0; i < n; ++i) {
		ss.erase({a[i], i});
		cout << ss.find_by_order((n - 1) / 2)->first << " ";
		ss.insert({a[i], i});
	}
	return 0;
}