output_description: If you will win, print `first`; if Lunlun will win, print `second`.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
      a_1
      a_2
      :
      a_N
attempt: #!/usr/bin/env python3
  
  import sys
  from typing import (
      Any,
      Callable,
      Deque,
      Dict,
      List,
      Mapping,
      Optional,
      Sequence,
      Set,
      Tuple,
      TypeVar,
      Union,
  )
  
  # import time
  # import math
  # import numpy as np
  # import scipy.sparse.csgraph as cs            # csgraph_from_dense(ndarray, null_value=inf), bellman_ford(G, return_predecessors=True), dijkstra, floyd_warshall
  # import random                                # random, uniform, randint, randrange, shuffle, sample
  # import string                                # ascii_lowercase, ascii_uppercase, ascii_letters, digits, hexdigits
  # import re                                    # re.compile(pattern) => ptn obj; p.search(s), p.match(s), p.finditer(s) => match obj; p.sub(after, s)
  # from bisect import bisect_left, bisect_right # bisect_left(a, x, lo=0, hi=len(a)) returns i such that all(val<x for val in a[lo:i]) and all(val>-=x for val in a[i:hi]).
  # from collections import deque                # deque class. deque(L): dq.append(x), dq.appendleft(x), dq.pop(), dq.popleft(), dq.rotate()
  # from collections import defaultdict          # subclass of dict. defaultdict(facroty)
  # from collections import Counter              # subclass of dict. Counter(iter): c.elements(), c.most_common(n), c.subtract(iter)
  # from datetime import date, datetime          # date.today(), date(year,month,day) => date obj; datetime.now(), datetime(year,month,day,hour,second,microsecond) => datetime obj; subtraction => timedelta obj
  # from datetime.datetime import strptime       # strptime('2019/01/01 10:05:20', '%Y/%m/%d/ %H:%M:%S') returns datetime obj
  # from datetime import timedelta               # td.days, td.seconds, td.microseconds, td.total_seconds(). abs function is also available.
  # from copy import copy, deepcopy              # use deepcopy to copy multi-dimentional matrix without reference
  # from functools import reduce                 # reduce(f, iter[, init])
  # from functools import lru_cache              # @lrucache ...arguments of functions should be able to be keys of dict (e.g. list is not allowed)
  # from heapq import heapify, heappush, heappop # built-in list. heapify(L) changes list in-place to min-heap in O(n), heappush(heapL, x) and heappop(heapL) in O(lgn).
  # from heapq import nlargest, nsmallest        # nlargest(n, iter[, key]) returns k-largest-list in O(n+klgn).
  # from itertools import count, cycle, repeat   # count(start[,step]), cycle(iter), repeat(elm[,n])
  # from itertools import groupby                # [(k, list(g)) for k, g in groupby('000112')] returns [('0',['0','0','0']), ('1',['1','1']), ('2',['2'])]
  # from itertools import starmap                # starmap(pow, [[2,5], [3,2]]) returns [32, 9]
  # from itertools import product, permutations  # product(iter, repeat=n), permutations(iter[,r])
  # from itertools import combinations, combinations_with_replacement
  # from itertools import accumulate             # accumulate(iter[, f])
  # from operator import itemgetter              # itemgetter(1), itemgetter('key')
  # from fractions import Fraction               # Fraction(a, b) => a / b ∈ Q. note: Fraction(0.1) do not returns Fraciton(1, 10). Fraction('0.1') returns Fraction(1, 10)
  
  
  def main():
      mod = 1000000007  # 10^9+7
      inf = float("inf")  # sys.float_info.max = 1.79e+308
      # inf = 2 ** 63 - 1             # (for fast JIT compile in PyPy) 9.22e+18
      sys.setrecursionlimit(10**6)  # 1000 -> 1000000
  
      def input():
          return sys.stdin.readline().rstrip()
  
      def ii():
          return int(input())
  
      def isp():
          return input().split()
  
      def mi():
          return map(int, input().split())
  
      def mi_0():
          return map(lambda x: int(x) - 1, input().split())
  
      def lmi():
          return list(map(int, input().split()))
  
      def lmi_0():
          return list(map(lambda x: int(x) - 1, input().split()))
  
      def li():
          return list(input())
  
      n = ii()
      L = [ii() for _ in range(n)]
      print("first") if any(map(lambda x: x % 2 == 1, L)) else print("second")
  
  
  if __name__ == "__main__":
      main()
  
problem_description: Statement
  
  There is an apple tree that bears apples of N colors. The N colors of these
  apples are numbered 1 to N, and there are a_i apples of Color i.
  
  You and Lunlun the dachshund alternately perform the following operation
  (starting from you):
  
    * Choose one or more apples from the tree and eat them. Here, the apples chosen at the same time must all have different colors.
  
  The one who eats the last apple from the tree will be declared winner. If both
  you and Lunlun play optimally, which will win?
samples: [{"input": "2\n    1\n    2", "output": "first\n    \n\nLet Color 1 be red, and Color 2 be blue. In this case, the tree bears one red\napple and two blue apples.\n\nYou should eat the red apple in your first turn. Lunlun is then forced to eat\none of the blue apples, and you can win by eating the other in your next turn.\n\nNote that you are also allowed to eat two apples in your first turn, one red\nand one blue (not a winning move, though).\n\n* * *"}, {"input": "3\n    100000\n    30000\n    20000", "output": "second"}]