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)

// https://atcoder.jp/contests/abc094/tasks/arc095_a

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <iostream>
#include <regex>
#include <set>
#include <tuple>
#include <vector>
using namespace std;

typedef long long ll;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;

#define EPS (1e-7)
#define INF (1e9)
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define REP(i, n) for (int i = 0; i < (n); ++i)
#define REPR(i, n) for (int i = n; i >= 0; i--)
#define FOREACH(x, a) for (auto &(x) : (a))
#define ALL(obj) (obj).begin(), (obj).end()
#define ALLR(obj) (obj).rbegin(), (obj).rend()

// long gcd(long a, long b) { return b ? gcd(b, a % b) : a; }
// long lcm(long a, long b) { return a * b / gcd(a, b); }

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin >> n;
    vector<int> v, w;
    v.resize(n);
    w.resize(n);
    for (int i = 0; i < n; i++) {
        // scanf("%d", v[i]);
        cin >> v[i];
    }
    copy(v.begin(), v.end(), w.begin());
    sort(ALL(w));

    for (int i = 0; i < n; i++) {
        if (v[i] < w[n / 2]) {
            printf("%d¥n", w[n / 2]);
        } else {
            printf("%d¥n", w[n / 2 - 1]);
        }
    }

    return 0;
}