output_description: Print the answer.
  
  * * *
input_description: The input is given from Standard Input in the following format:
  
      
      
      N
      A_1 A_2 A_3 ... A_{N}
attempt: import sys
  import math
  import collections
  import itertools
  import array
  import inspect
  
  # Set max recursion limit
  sys.setrecursionlimit(1000000)
  
  
  # Debug output
  def chkprint(*args):
      names = {id(v): k for k, v in inspect.currentframe().f_back.f_locals.items()}
      print(", ".join(names.get(id(arg), "???") + " = " + repr(arg) for arg in args))
  
  
  # Binary converter
  def to_bin(x):
      return bin(x)[2:]
  
  
  def li_input():
      return [int(_) for _ in input().split()]
  
  
  def gcd(n, m):
      if n % m == 0:
          return m
      else:
          return gcd(m, n % m)
  
  
  def gcd_list(L):
      v = L[0]
  
      for i in range(1, len(L)):
          v = gcd(v, L[i])
  
      return v
  
  
  def lcm(n, m):
      return (n * m) // gcd(n, m)
  
  
  def lcm_list(L):
      v = L[0]
  
      for i in range(1, len(L)):
          v = lcm(v, L[i])
  
      return v
  
  
  # Width First Search (+ Distance)
  def wfs_d(D, N, K):
      """
      D: 隣接行列(距離付き)
      N: ノード数
      K: 始点ノード
      """
  
      dfk = [-1] * (N + 1)
      dfk[K] = 0
  
      cps = [(K, 0)]
      r = [False] * (N + 1)
      r[K] = True
      while len(cps) != 0:
          n_cps = []
          for cp, cd in cps:
              for i, dfcp in enumerate(D[cp]):
                  if dfcp != -1 and not r[i]:
                      dfk[i] = cd + dfcp
                      n_cps.append((i, cd + dfcp))
                      r[i] = True
  
          cps = n_cps[:]
  
      return dfk
  
  
  # Depth First Search (+Distance)
  def dfs_d(v, pre, dist):
      """
      v:  現在のノード
      pre: １つ前のノード
      dist: 現在の距離
  
      以下は別途用意する
      D: 隣接リスト(行列ではない)
      D_dfs_d: dfs_d関数で用いる，始点ノードから見た距離リスト
      """
  
      global D
      global D_dfs_d
  
      D_dfs_d[v] = dist
  
      for next_v, d in D[v]:
          if next_v != pre:
              dfs_d(next_v, v, dist + d)
  
      return
  
  
  def sigma(N):
      ans = 0
      for i in range(1, N + 1):
          ans += i
      return ans
  
  
  def comb(n, r):
      if n - r < r:
          r = n - r
      if r == 0:
          return 1
      if r == 1:
          return n
  
      numerator = [n - r + k + 1 for k in range(r)]
      denominator = [k + 1 for k in range(r)]
  
      for p in range(2, r + 1):
          pivot = denominator[p - 1]
          if pivot > 1:
              offset = (n - r) % p
              for k in range(p - 1, r, p):
                  numerator[k - offset] /= pivot
                  denominator[k] /= pivot
  
      result = 1
      for k in range(r):
          if numerator[k] > 1:
              result *= int(numerator[k])
  
      return result
  
  
  def bisearch(L, target):
      low = 0
      high = len(L) - 1
  
      while low <= high:
          mid = (low + high) // 2
          guess = L[mid]
          if guess == target:
              return True
          elif guess < target:
              low = mid + 1
          elif guess > target:
              high = mid - 1
      if guess != target:
          return False
  
  
  # --------------------------------------------
  
  dp = None
  
  
  def main():
      N = int(input())
      A = set(li_input())
  
      if len(A) % 2 == 0:
          print(len(A) - 1)
      else:
          print(len(A))
  
  
  main()
  
problem_description: Statement
  
  Snuke has decided to play a game using cards. He has a deck consisting of N
  cards. On the i-th card from the top, an integer A_i is written.
  
  He will perform the operation described below zero or more times, so that the
  values written on the remaining cards will be pairwise distinct. Find the
  maximum possible number of remaining cards. Here, N is odd, which guarantees
  that at least one card can be kept.
  
  Operation: Take out three arbitrary cards from the deck. Among those three
  cards, eat two: one with the largest value, and another with the smallest
  value. Then, return the remaining one card to the deck.
samples: [{"input": "5\n    1 2 1 3 7", "output": "3\n    \n\nOne optimal solution is to perform the operation once, taking out two cards\nwith 1 and one card with 2. One card with 1 and another with 2 will be eaten,\nand the remaining card with 1 will be returned to deck. Then, the values\nwritten on the remaining cards in the deck will be pairwise distinct: 1, 3 and\n7.\n\n* * *"}, {"input": "15\n    1 3 5 2 1 3 2 8 8 6 2 6 11 1 1", "output": "7"}]