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)

// May this submission get accepted!

#pragma GCC optimize ("O3")
#pragma GCC target ("tune=native")
#pragma GCC target ("avx")

#include <bits/stdc++.h>
#include <unordered_map>

// 汎用マクロ
#define ALL_OF(x) (x).begin(), (x).end()
#define REP(i,n) for (long long i=0, i##_len=(n); i<i##_len; i++)
#define RANGE(i,is,ie) for (long long i=(is), i##_end=(ie); i<=i##_end; i++)
#define DSRNG(i,is,ie) for (long long i=(is), i##_end=(ie); i>=i##_end; i--)
#define STEP(i, is, ie, step) for (long long i=(is), i##_end=(ie), i##_step = (step); i<=i##_end; i+=i##_step)
#define UNIQUE(v) do { sort((v).begin(), (v).end()); (v).erase(unique((v).begin(), (v).end()), (v).end()); } while (false)
template<class T> bool chmax(T &a, const T &b) { if (a < b) {a = b; return true;} return false; }
template<class T> bool chmin(T &a, const T &b) { if (a > b) {a = b; return true;} return false; }
#define INF 0x7FFFFFFF
#define LINF 0x7FFFFFFFFFFFFFFFLL
#define Yes(q) ((q) ? "Yes" : "No")
#define YES(q) ((q) ? "YES" : "NO")
#define Possible(q) ((q) ? "Possible" : "Impossible")
#define POSSIBLE(q) ((q) ? "POSSIBLE" : "IMPOSSIBLE")
#define DUMP(q) cerr << "[DEBUG] " #q ": " << (q) << " at " __FILE__ ":" << __LINE__ << endl
#define DUMPALL(q) do { cerr << "[DEBUG] " #q ": ["; REP(DUMPALL_ITR, (q).size()) { cerr << (q)[DUMPALL_ITR] << (DUMPALL_ITR == DUMPALL_ITR_len-1 ? "" : ", "); } cerr << "] at " __FILE__ ":" << __LINE__ << endl; } while (false)
template<class T> T gcd(T a, T b) { if (a < b) std::swap(a, b); while (b) std::swap(a %= b, b); return a; }
template<class T> T lcm(const T a, const T b) { return a / gcd(a, b) * b; }

// gcc拡張マクロ
#define popcount __builtin_popcount
#define popcountll __builtin_popcountll

// エイリアス
#define DANCE_ long
#define ROBOT_ unsigned
#define HUMAN_ signed
using  ll = DANCE_ HUMAN_ DANCE_;
using ull = DANCE_ ROBOT_ DANCE_;
using  ld = long double;
using namespace std;

// 標準入出力
struct inp {
    size_t sz;
    inp(size_t _sz = 1) : sz(_sz) {}
    template <typename T> operator T () const { T a; cin >> a; return a; }
    template <typename T> operator vector<T> () const { vector<T> a(sz); for (size_t i = 0; i < sz; i++) cin >> a[i]; return a; }
    template <typename T, typename U> operator pair<T, U> () const { T f; U s; cin >> f >> s; return pair<T, U>(f, s); }
};
inp inp1; // input one
template <typename T> void say(T &x, char end = '\n') { cout << x << end; }
template <typename T> void say(vector<T> &x, char sep = ' ', char end = '\n') { for (size_t i = 0, sz = x.size(); i < sz; i++) { cout << x[i] << (i == sz-1 ? end : sep); } }

// モジュール


// 処理内容
int main() {
    
    ios::sync_with_stdio(false); // stdioを使うときはコメントアウトすること
    cin.tie(nullptr);            // インタラクティブ問題ではコメントアウトすること

    ll n = inp1;
    vector<ll> x = inp(n);
    auto y = x;
    sort(ALL_OF(y));
    ll xmid = y[n/2], xmed = y[n/2-1];

    REP(i, n) {
        if (x[i] < xmid) {
            cout << xmid << endl;
        } else {
            cout << xmed << endl;
        }
    }
    
}