output_description: Print N lines. The i-th line should contain B_i.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
      X_1 X_2 ... X_N
attempt: #include<bits/stdc++.h>
  using namespace std;
  typedef long long ll;
  typedef vector<ll> vi;
  typedef pair<ll,ll> pi;
  typedef pair<ll,pi> pii;
  typedef vector<pi> vpi;
  typedef set<ll> si;
  typedef long double ld;
  #define mp make_pair
  #define pb push_back
  #define f first
  #define s second
  ll INF = 1e18;
  ll MOD = 1e9+7;
  #define lb lower_bound
  #define ub upper_bound
  #define SZ(x) (ll)x.size()
  #define ALL(x) x.begin(),x.end()
  int l,r,A[100100], B[100100], N;
  
  int main(){
    cin>>N;
    for (int i=0;i<N;++i){cin>>A[i];B[i] = A[i];}
    sort(A,A+N);
    l = A[N/2-1];
    r = A[N/2];
    for (int i=0;i<N;++i){
      if (B[i] <= l)cout<<r<<'\n';
      else cout<<l<<'\n';
    }
  }
problem_description: 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.
samples: [{"input": "4\n    2 4 4 3", "output": "4\n    3\n    3\n    4\n    \n\n  * Since the median of X_2, X_3, X_4 is 4, B_1 = 4.\n  * Since the median of X_1, X_3, X_4 is 3, B_2 = 3.\n  * Since the median of X_1, X_2, X_4 is 3, B_3 = 3.\n  * Since the median of X_1, X_2, X_3 is 4, B_4 = 4.\n\n* * *"}, {"input": "2\n    1 2", "output": "2\n    1\n    \n\n* * *"}, {"input": "6\n    5 5 4 4 3 3", "output": "4\n    4\n    4\n    4\n    4\n    4"}]