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: ###############################################################################
  
  from sys import stdout
  from bisect import bisect_left as binl
  from copy import copy, deepcopy
  from collections import defaultdict
  
  
  mod = 1
  
  
  def intin():
      input_tuple = input().split()
      if len(input_tuple) <= 1:
          return int(input_tuple[0])
      return tuple(map(int, input_tuple))
  
  
  def intina():
      return [int(i) for i in input().split()]
  
  
  def intinl(count):
      return [intin() for _ in range(count)]
  
  
  def modadd(x, y):
      global mod
      return (x + y) % mod
  
  
  def modmlt(x, y):
      global mod
      return (x * y) % mod
  
  
  def lcm(x, y):
      while y != 0:
          z = x % y
          x = y
          y = z
      return x
  
  
  def combination(x, y):
      assert x >= y
  
      if y > x // 2:
          y = x - y
  
      ret = 1
  
      for i in range(0, y):
          j = x - i
          i = i + 1
  
          ret = ret * j
          ret = ret // i
  
      return ret
  
  
  def get_divisors(x):
      retlist = []
      for i in range(1, int(x**0.5) + 3):
          if x % i == 0:
              retlist.append(i)
              retlist.append(x // i)
      return retlist
  
  
  def get_factors(x):
      retlist = []
      for i in range(2, int(x**0.5) + 3):
          while x % i == 0:
              retlist.append(i)
              x = x // i
      retlist.append(x)
      return retlist
  
  
  def make_linklist(xylist):
      linklist = {}
      for a, b in xylist:
          linklist.setdefault(a, [])
          linklist.setdefault(b, [])
          linklist[a].append(b)
          linklist[b].append(a)
      return linklist
  
  
  def calc_longest_distance(linklist, v=1):
      distance_list = {}
      distance_count = 0
      distance = 0
      vlist_previous = []
      vlist = [v]
      nodecount = len(linklist)
  
      while distance_count < nodecount:
          vlist_next = []
          for v in vlist:
              distance_list[v] = distance
              distance_count += 1
              vlist_next.extend(linklist[v])
          distance += 1
          vlist_to_del = vlist_previous
          vlist_previous = vlist
          vlist = list(set(vlist_next) - set(vlist_to_del))
  
      max_distance = -1
      max_v = None
      for v, distance in distance_list.items():
          if distance > max_distance:
              max_distance = distance
              max_v = v
  
      return (max_distance, max_v)
  
  
  def calc_tree_diameter(linklist, v=1):
      _, u = calc_longest_distance(linklist, v)
      distance, _ = calc_longest_distance(linklist, u)
      return distance
  
  
  class UnionFind:
  
      def __init__(self, n):
          self.parent = [i for i in range(n)]
  
      def root(self, i):
          if self.parent[i] == i:
              return i
          self.parent[i] = self.root(self.parent[i])
          return self.parent[i]
  
      def unite(self, i, j):
          rooti = self.root(i)
          rootj = self.root(j)
          if rooti == rootj:
              return
          if rooti < rootj:
              self.parent[rootj] = rooti
          else:
              self.parent[rooti] = rootj
  
      def same(self, i, j):
          return self.root(i) == self.root(j)
  
  
  ###############################################################################
  
  
  def main():
      n = intin()
      xlist = intina()
  
      xxlist = sorted(xlist)
      lxx = len(xxlist)
      c1 = xxlist[lxx // 2 - 1]
      c2 = xxlist[lxx // 2]
  
      for x in xlist:
          if x <= c1:
              print(c2)
          else:
              print(c1)
  
  
  if __name__ == "__main__":
      main()
  
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"}]