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)

#include<bits/stdc++.h>
using namespace std;
struct C{
int id,val;
};
deque<C> mergesort(deque<C> x){
    deque<C> sorted;
    if(x.size()==1)
        return x;
    int high=x.size(),low=0;
    int mid=low+high>>2;
    deque <C> l,r;
    for(int i=0;i<mid;i++){
        l.push_back(x[i]);
    }
    for(int i=mid;i<high;i++){
        r.push_back(x[i]);
    }
    l=mergesort(l);
    r=mergesort(r);

    while(!l.empty()&&!r.empty()){
        if(l.front().val<r.front().val){
            sorted.push_back(l.front());
            l.pop_front();
        }
        else{
           sorted.push_back(r.front());
           r.pop_front();
        }
    }
    while(!l.empty()){
        sorted.push_back(l.front());
        l.pop_front();
    }
    while(!r.empty()){
        sorted.push_back(r.front());
        r.pop_front();
    }
    return sorted;
}
int main(){
    int n,x;
    scanf("%d",&n);
    deque<C> a;
    //C a[n+1],dd;
    for(int i=1;i<=n;i++){
        scanf("%d",&x);
        a.push_back({i,x});
    }
    for(int i=1;i<=n;i++){
            int c=0;
        for(int j=1;j<=n/2;j++){
            if(i==a[j].id){
                printf("%d\n",a[n/2+1].val);
                c++;
                break;
            }
        }
        if(c==0){
            printf("%d\n",a[n/2].val);
        }
    }

}