output_description: If Fennec wins, print `Fennec`; if Snuke wins, print `Snuke`.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
      a_1 b_1
      :
      a_{N-1} b_{N-1}
attempt: # -*- coding: utf-8 -*-
  
  #############
  # Libraries #
  #############
  
  import sys
  
  input = sys.stdin.readline
  
  import math
  
  # from math import gcd
  import bisect
  import heapq
  from collections import defaultdict
  from collections import deque
  from collections import Counter
  from functools import lru_cache
  
  #############
  # Constants #
  #############
  
  MOD = 10**9 + 7
  INF = float("inf")
  AZ = "abcdefghijklmnopqrstuvwxyz"
  
  #############
  # Functions #
  #############
  
  
  ######INPUT######
  def I():
      return int(input().strip())
  
  
  def S():
      return input().strip()
  
  
  def IL():
      return list(map(int, input().split()))
  
  
  def SL():
      return list(map(str, input().split()))
  
  
  def ILs(n):
      return list(int(input()) for _ in range(n))
  
  
  def SLs(n):
      return list(input().strip() for _ in range(n))
  
  
  def ILL(n):
      return [list(map(int, input().split())) for _ in range(n)]
  
  
  def SLL(n):
      return [list(map(str, input().split())) for _ in range(n)]
  
  
  ######OUTPUT######
  def P(arg):
      print(arg)
      return
  
  
  def Y():
      print("Yes")
      return
  
  
  def N():
      print("No")
      return
  
  
  def E():
      exit()
  
  
  def PE(arg):
      print(arg)
      exit()
  
  
  def YE():
      print("Yes")
      exit()
  
  
  def NE():
      print("No")
      exit()
  
  
  #####Shorten#####
  def DD(arg):
      return defaultdict(arg)
  
  
  #####Inverse#####
  def inv(n):
      return pow(n, MOD - 2, MOD)
  
  
  ######Combination######
  kaijo_memo = []
  
  
  def kaijo(n):
      if len(kaijo_memo) > n:
          return kaijo_memo[n]
      if len(kaijo_memo) == 0:
          kaijo_memo.append(1)
      while len(kaijo_memo) <= n:
          kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD)
      return kaijo_memo[n]
  
  
  gyaku_kaijo_memo = []
  
  
  def gyaku_kaijo(n):
      if len(gyaku_kaijo_memo) > n:
          return gyaku_kaijo_memo[n]
      if len(gyaku_kaijo_memo) == 0:
          gyaku_kaijo_memo.append(1)
      while len(gyaku_kaijo_memo) <= n:
          gyaku_kaijo_memo.append(
              gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo), MOD - 2, MOD) % MOD
          )
      return gyaku_kaijo_memo[n]
  
  
  def nCr(n, r):
      if n == r:
          return 1
      if n < r or r < 0:
          return 0
      ret = 1
      ret = ret * kaijo(n) % MOD
      ret = ret * gyaku_kaijo(r) % MOD
      ret = ret * gyaku_kaijo(n - r) % MOD
      return ret
  
  
  ######Factorization######
  def factorization(n):
      arr = []
      temp = n
      for i in range(2, int(-(-(n**0.5) // 1)) + 1):
          if temp % i == 0:
              cnt = 0
              while temp % i == 0:
                  cnt += 1
                  temp //= i
              arr.append([i, cnt])
      if temp != 1:
          arr.append([temp, 1])
      if arr == []:
          arr.append([n, 1])
      return arr
  
  
  #####MakeDivisors######
  def make_divisors(n):
      divisors = []
      for i in range(1, int(n**0.5) + 1):
          if n % i == 0:
              divisors.append(i)
              if i != n // i:
                  divisors.append(n // i)
      return divisors
  
  
  #####MakePrimes######
  def make_primes(N):
      max = int(math.sqrt(N))
      seachList = [i for i in range(2, N + 1)]
      primeNum = []
      while seachList[0] <= max:
          primeNum.append(seachList[0])
          tmp = seachList[0]
          seachList = [i for i in seachList if i % tmp != 0]
      primeNum.extend(seachList)
      return primeNum
  
  
  #####GCD#####
  def gcd(a, b):
      while b:
          a, b = b, a % b
      return a
  
  
  #####LCM#####
  def lcm(a, b):
      return a * b // gcd(a, b)
  
  
  #####BitCount#####
  def count_bit(n):
      count = 0
      while n:
          n &= n - 1
          count += 1
      return count
  
  
  #####ChangeBase#####
  def base_10_to_n(X, n):
      if X // n:
          return base_10_to_n(X // n, n) + [X % n]
      return [X % n]
  
  
  def base_n_to_10(X, n):
      return sum(int(str(X)[-i - 1]) * n**i for i in range(len(str(X))))
  
  
  def base_10_to_n_without_0(X, n):
      X -= 1
      if X // n:
          return base_10_to_n_without_0(X // n, n) + [X % n]
      return [X % n]
  
  
  #####IntLog#####
  def int_log(n, a):
      count = 0
      while n >= a:
          n //= a
          count += 1
      return count
  
  
  #############
  # Main Code #
  #############
  
  N = I()
  graph = [[] for i in range(N)]
  for _ in range(N - 1):
      a, b = IL()
      graph[a - 1].append(b - 1)
      graph[b - 1].append(a - 1)
  
  
  s = 0
  
  d1 = [INF] * N
  d1[s] = 0
  q = deque([s])
  while q:
      u = q.popleft()
      for v in graph[u]:
          if d1[v] != INF:
              continue
          d1[v] = d1[u] + 1
          q.append(v)
  
  s = N - 1
  d2 = [INF] * N
  d2[s] = 0
  q = deque([s])
  while q:
      u = q.popleft()
      for v in graph[u]:
          if d2[v] != INF:
              continue
          d2[v] = d2[u] + 1
          q.append(v)
  
  ans = 0
  for i in range(N):
      if d1[i] <= d2[i]:
          ans += 1
  
  if ans >= (N + 1) // 2:
      print("Fennec")
  else:
      print("Snuke")
  
problem_description: Statement
  
  Fennec and Snuke are playing a board game.
  
  On the board, there are N cells numbered 1 through N, and N-1 roads, each
  connecting two cells. Cell a_i is adjacent to Cell b_i through the i-th road.
  Every cell can be reached from every other cell by repeatedly traveling to an
  adjacent cell. In terms of graph theory, the graph formed by the cells and the
  roads is a tree.
  
  Initially, Cell 1 is painted black, and Cell N is painted white. The other
  cells are not yet colored. Fennec (who goes first) and Snuke (who goes second)
  alternately paint an uncolored cell. More specifically, each player performs
  the following action in her/his turn:
  
    * Fennec: selects an uncolored cell that is adjacent to a **black** cell, and paints it **black**.
    * Snuke: selects an uncolored cell that is adjacent to a **white** cell, and paints it **white**.
  
  A player loses when she/he cannot paint a cell. Determine the winner of the
  game when Fennec and Snuke play optimally.
samples: [{"input": "7\n    3 6\n    1 2\n    3 1\n    7 4\n    5 7\n    1 4", "output": "Fennec\n    \n\nFor example, if Fennec first paints Cell 2 black, she will win regardless of\nSnuke's moves.\n\n* * *"}, {"input": "4\n    1 4\n    4 2\n    2 3", "output": "Snuke"}]