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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>

#define MAX 500000
#define SENTINEL 200000000

int L[MAX/2+2],R[MAX/2+2];
//int cnt=0;  cnt=nlogn

void merge(int A[],int n,int left,int mid,int right){ //分割と統治
    int n1 = mid-left;
    int n2 = right-mid;
    for(int i=0;i<n1;i++) L[i]=A[left+i];
    for(int i=0;i<n2;i++) R[i]=A[mid+i];
    L[n1]=R[n2]=SENTINEL;  //番兵
    int i=0,j=0;
    for(int k=left;k<right;k++){
        //cnt++;
        if(L[i]<=R[j]){
            A[k]=L[i++];
        }else{
            A[k]=R[j++];
        }
    }
}

void mergeSort(int A[],int n,int left,int right){
    if(left+1<right){  //要素数が2つ以上
        int mid = (left+right)/2;
        mergeSort(A,n,left,mid);
        mergeSort(A,n,mid,right);
        merge(A,n,left,mid,right);
    }
}

int main(void){
    int n,i;
    scanf("%d",&n);
    int x[n],b[n];
    for(i=0;i<n;i++){
        scanf("%d",&x[i]);
        b[i] = x[i];
    }
    
    mergeSort(b,n,0,n);
    
    for(i=0;i<n;i++){
        if(x[i]<b[n/2]){
            printf("%d\n",b[n/2]);
        }else{
            printf("%d\n",b[(n/2)-1]);
        }
    }
    return 0;
}