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)

#include <bits/stdc++.h>

using namespace std;

template <typename T> T gcd(T x, T y) {
  if (y == 0)
    return x;
  return gcd(y, x % y);
}

template <typename T> T lcm(T x, T y) {
  if (x == 0 || y == 0)
    return 0;
  return x / gcd(x, y) * y;
}

int main() {
  int N;
  cin >> N;
  vector<int> X(N);
  for (int i = 0; i < N; i++) {
    cin >> X[i];
  }
  vector<int> A = X;
  sort(A.begin(), A.end());
  int mid1 = A[(N - 1) / 2];
  int mid2 = A[N / 2];
  for (int i = 0; i < N; i++) {
    if (mid1 < X[i])
      cout << mid1 << endl;
    else
      cout << mid2 << endl;
  }
}