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 Kotlin (1.0.0) (judge verdict: Accepted)

fun main(args: Array<String>) {
    val n = readLine()?.toInt() ?: return
    val xList = readLine()?.split(" ")?.map { it.toInt() } ?: return

    val middleIndex = xList.size / 2
    val sortedX = xList.sorted()

    val middleValue = sortedX[middleIndex] // 真っ二つにしたときに、大きい方の1番目に該当する値
    val lessMiddleValue = sortedX[middleIndex - 1] // 真っ二つにしたときに、小さい方のlastに該当する値

    // Xiが↑の中央値より大きいかどうか
    val moreOrLessList = xList.withIndex().map { it.value >= middleValue }

    for (isMore in moreOrLessList) {
        if (isMore) {
            println(lessMiddleValue)
        } else {
            println(middleValue)
        }
    }
}