Problem:


Score : 400 points

Problem 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.

Constraints

2 \leq N \leq 10^5

1 \leq a_i, b_i \leq N

The given graph is a tree.

Input

Input is given from Standard Input in the following format:

N
a_1 b_1
:
a_{N-1} b_{N-1}

Output

If Fennec wins, print Fennec; if Snuke wins, print Snuke.

Sample Input 1

7
3 6
1 2
3 1
7 4
5 7
1 4

Sample Output 1

Fennec

For example, if Fennec first paints Cell 2 black, she will win regardless of Snuke's moves.

Sample Input 2

4
1 4
4 2
2 3

Sample Output 2

Snuke



Sample input:
7
3 6
1 2
3 1
7 4
5 7
1 4


Sample output:
Fennec

Submission status: Runtime Error

Solution code:
n = int(input())
edges = [[] for i in range(n)]
for i in range(n-1):
    a,b = map(int, input().split())
    a -= 1
    b -= 1
    edges[a].append(b)
    edges[b].append(a)
f = 0
s = 0
from collections import deque
ps = [-1] * n
used = [False] * n
used[0] = True # 始めどこから行くか
q = deque([0])
while len(q) > 0:
    a = q.popleft()
    if a == n-1:
        break
    for u in edges[a]:
        if not used[u]:
            ps[u] = a
            q.append(u)
            used[u] = True
nx = n - 1
xs = [n-1]
while nx != 0:
    nx = ps[nx]
    xs.append(nx)
l = len(xs) - 2
f = (l + 1)//2
s = l - f

mid = len(xs)//2

# bfs 0側
used = [-1] * n
used[0] = 0 
q = deque([0])
cnt = 0
while len(q) > 0:
    a = q.popleft()
    for u in edges[a]:
        if used[u] == -1 and u != xs[mid+1]:
            q.append(u)
            used[u] = 0
            cnt += 1
# print("cnt",cnt)
f += cnt

# bfs n-1側
used = [-1] * n
used[n-1] = 0 
q = deque([n-1])
cnt = 0
while len(q) > 0:
    a = q.popleft()
    for u in edges[a]:
        if used[u] == -1 and u != xs[mid]:
            q.append(u)
            used[u] = 0
            cnt += 1
# print("cnt",cnt)
s += cnt

if f > s:
    print("Fennec")
else:
    print("Snuke")

Submission status: Runtime Error

Solution code:
n = int(input())
edges = [[] for _ in range(n)]

for _ in range(n-1):
    a, b = map(int, input().split())
    a -= 1
    b -= 1
    edges[a].append(b)
    edges[b].append(a)

seen = [-1] * n


def dfs(node, prev):
    if seen[node] != -1:
        return
    seen[node] = prev

    for to in edges[node]:
        dfs(to, node)


dfs(0, 0)

colors = [0] * n

node = n-1
goal_to_start = [n-1]
while node:
    node = seen[node]
    goal_to_start.append(node)

length = len(goal_to_start)
limit = (length) // 2

colors = [0] * n
for node in goal_to_start[:limit]:
    colors[node] = 2

for node in goal_to_start[limit:]:
    colors[node] = 1

fennec_todo = [node for node in range(n) if colors[node] == 1]
snuke_todo = [node for node in range(n) if colors[node] == 2]


def paint(todo, my_color):
    while todo:
        node = todo.pop()
        for to in edges[node]:
            if colors[to]:
                continue

            colors[to] = my_color
            todo.append(to)


paint(fennec_todo, 1)
paint(snuke_todo, 2)

feccec_count = len([color for color in colors if color == 1])
snuke_count = len([color for color in colors if color == 2])

if feccec_count > snuke_count:
    print('Fennec ')

else:
    print('Snuke ')

Submission status: Runtime Error

Solution code:
from collections import deque

N=int(input())
graph=[[] for _ in range(N)]
for _ in range(N-1):
    a,b=map(int,input().split())
    a-=1
    b-=1
    graph[a].append(b)
    graph[b].append(a)


dist=[float("inf")]*N
dist[0]=0
prev=[None]*N

def dfs1(v,parent,d):
    prev[v]=parent
    dist[v]=d
    for v2 in graph[v]:
        if v2==parent:
            continue
        dfs1(v2,v,d+1)

dfs1(0,-1,0)

x=N-1
for _ in range((dist[-1]-1)//2):
    x=prev[x]
cant_go=prev[x]

sunuke_score=0
def dfs2(v,parent):
    global sunuke_score
    sunuke_score+=1
    for v2 in graph[v]:
        if v2==parent:
            continue
        if v2==cant_go:
            continue
        dfs2(v2,v)

dfs2(x,-1)

fennec_score=N-sunuke_score
if sunuke_score>fennec_score:
    print("Snuke")
else:
    print("Fennec")

Submission status: Runtime Error

Solution code:
# 2020-05-28 22:45:55

import sys
sys.setrecursionlimit(10**8)
input = sys.stdin.readline

N = int(input())
ab = [list(map(int,input().split())) for _ in range (N-1)]

edge = [[] for _ in range(N)]
for u,v in ab:
    edge[u-1].append(v-1)
    edge[v-1].append(u-1)

fromNode = [0] * N
def dfs(v, p):
    for nv in edge[v]:
        if nv==p:continue 
        fromNode[nv] = v
        dfs(nv, v)
    return 

def pathFind(s, g, fromNode):
    path = []
    while s!=g:
        g = fromNode[g]
        if g==s:break
        path.append(g)
    path.reverse()
    return path
dfs(0,-1)
path = pathFind(0,N-1,fromNode)

colors = [-1]*N
for i in range(len(path)):
    if i%2==0:
        colors[path[i//2]] = 0
    else:
        colors[path[N-1-(i//2)]] = 1

def dfs_color(v, p, c):
    for nv in edge[v]:
        if nv==p:continue 
        if colors[nv]==[-1]:
            colors[nv] = c
            dfs_color(nv, v, c)
    return 

dfs_color(0,-1,0)
dfs_color(N-1, -1, 1)
for p in path:
    c = colors[p]
    dfs_color(p, -1, c)

B = 0
W = 0
for c in colors:
    if c==0:B+=1
    elif c==1:W+=1

if B>W:
    print('Fennec')
else:
    print('Snuke')

Submission status: Runtime Error

Solution code:
# 2020-05-28 22:45:55

import sys
sys.setrecursionlimit(10**8)
input = sys.stdin.readline

N = int(input())
ab = [list(map(int,input().split())) for _ in range (N-1)]

edge = [[] for _ in range(N)]
for u,v in ab:
    edge[u-1].append(v-1)
    edge[v-1].append(u-1)

fromNode = [-1] * N
def dfs(v, p):
    for nv in edge[v]:
        if nv==p:continue 
        fromNode[nv] = v
        dfs(nv, v)
    return 

def pathFind(s, g, fromNode):
    path = []
    while s!=g:
        g = fromNode[g]
        if g==s:break
        path.append(g)
    path.reverse()
    return path
dfs(0,-1)
path = pathFind(0,N-1,fromNode)

colors = [-1]*N
for i in range(len(path)):
    if i%2==0:
        colors[path[i//2]] = 0
    else:
        colors[path[N-1-(i//2)]] = 1

def dfs_color(v, p, c):
    for nv in edge[v]:
        if nv==p:continue 
        if colors[nv]==[-1]:
            colors[nv] = c
            dfs_color(nv, v, c)
    return 

dfs_color(0,-1,0)
dfs_color(N-1, -1, 1)
for p in path:
    c = colors[p]
    dfs_color(p, -1, c)

B = 0
W = 0
for c in colors:
    if c==0:B+=1
    elif c==1:W+=1

if B>W:
    print('Fennec')
else:
    print('Snuke')

Submission status: Runtime Error

Solution code:
def main(sample_file = ''):

    """ convenient functions
    # for i, a in enumerate(iterable)
    # q, mod = divmod(a, b)
    # divmod(x, y) returns the tuple (x//y, x%y)
    # Higher-order function: reduce(operator.mul, xyz_count, 1)
    # manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
    """

    """convenient decorator
    # @functools.lru_cache():
    # to facilitate use of recursive function
        # ex:
        # from functools import lru_cache
        # import sys
        # sys.setrecursionlimit(10**9)
        # @lru_cache(maxsize=None)
        # def fib(n):
        #     if n < 2:
        #         return n
        #     return fib(n-1) + fib(n-2)
        # print(fib(1000))
    """
    
    # import numpy as np
    import sys
    sys.setrecursionlimit(10**7)
    from itertools import accumulate, combinations, permutations, product # https://docs.python.org/ja/3/library/itertools.html
    # accumulate() returns iterator! to get list: list(accumulate())
    from math import factorial, ceil, floor, sqrt
    def factorize(n):
        """return the factors of the Arg and count of each factor
        
        Args:
            n (long): number to be resolved into factors
        
        Returns:
            list of tuples: factorize(220) returns [(2, 2), (5, 1), (11, 1)]
        """
        fct = []  # prime factor
        b, e = 2, 0  # base, exponent
        while b * b <= n:
            while n % b == 0:
                n = n // b
                e = e + 1
            if e > 0:
                fct.append((b, e))
            b, e = b + 1, 0
        if n > 1:
            fct.append((n, 1))
        return fct
    def combinations_count(n, r):   
        """Return the number of selecting r pieces of items from n kinds of items.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        # TODO: How should I do when n - r is negative?
        if n < 0 or r < 0:
            raise Exception('combinations_count(n, r) not defined when n or r is negative')
        if n - r < r: r = n - r
        if r < 0: return 0
        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 combinations_with_replacement_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items allowing individual elements to be repeated more than once.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        if n < 0 or r < 0:
            raise Exception('combinations_with_replacement_count(n, r) not defined when n or r is negative')
        elif n == 0:
            return 1
        else:
            return combinations_count(n + r - 1, r)
    from bisect import bisect_left, bisect_right
    from collections import deque, Counter, defaultdict # https://docs.python.org/ja/3/library/collections.html#collections.deque
    from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
    from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
    import operator
    from operator import itemgetter #sort 
    # ex1: List.sort(key=itemgetter(1))
    # ex2: sorted(tuples, key=itemgetter(1,2))
    from functools import reduce, lru_cache
    def chmin(x, y):
        """change minimum
        if x > y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current minimum value
            y (long): potential minimum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x > y:
            x = y
            return (x, True)
        else:
            return (x, False)
    def chmax(x, y):
        """change maximum
        if x < y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current maximum value
            y (long): potential maximum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x < y:
            x = y
            return (x, True)
        else:
            return (x, False)

    from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
    def gcds(numbers):
        return reduce(gcd, numbers)
    def lcm(x, y):
        return (x * y) // gcd(x, y)
    def lcms(numbers):
        return reduce(lcm, numbers, 1)
    def make_divisors(n, reversed=False):
        """create list of divisors
        
        Args:
            number (int): number from which list of divisors is created
            reversed (bool, optional): ascending order if False. Defaults to False.
        
        Returns:
            list: list of divisors
        """
        divisors = set()
        for i in range(1, int(n**0.5)+1):
            if n % i == 0:
                divisors.add(i)
                divisors.add(n//i)
        return sorted(list(divisors),reverse=reversed)
    
    # first create factorial_list
    # fac_list = mod_factorial_list(n)
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    AtoZ = list('abcdefghijklmnopqrsruvwxyz') # 26
    modpow = lambda a, n, p = MOD: pow(a, n, p) # Recursive function in python is slow!
    def modinv(a, p = MOD):
        # evaluate reciprocal using Fermat's little theorem:
        # a**(p-1) is identical to 1 (mod p) when a and p is coprime
        return modpow(a, p-2, p)
    def modinv_list(n, p = MOD):
        if n <= 1:
            return [0,1][:n+1]
        else:
            inv_t = [0,1]
            for i in range(2, n+1):
                inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
            return inv_t
    def modfactorial_list(n, p = MOD):
        if n == 0:
            return [1]
        else:
            l = [0] * (n+1)
            tmp = 1
            for i in range(1, n+1):
                tmp = tmp * i % p
                l[i] = tmp
            return l
    def modcomb(n, k, fac_list = [], p = MOD):
        # fac_list = modfactorial_list(100)
        # print(modcomb(100, 5, modfactorial_list(100)))
        from math import factorial
        if n < 0 or k < 0 or n < k: return 0
        if n == 0 or k == 0: return 1
        if len(fac_list) <= n:
            a = factorial(n) % p
            b = factorial(k) % p
            c = factorial(n-k) % p
        else:
            a = fac_list[n]
            b = fac_list[k]
            c = fac_list[n-k]
        return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
    def modadd(a, b, p = MOD):
        return (a + b) % MOD
    def modsub(a, b, p = MOD):
        return (a - b) % p
    def modmul(a, b, p = MOD):
        return ((a % p) * (b % p)) % p
    def moddiv(a, b, p = MOD):
        return modmul(a, modpow(b, p-2, p))

    class UnionFindTree:
        """union find tree class
        TODO: fix this description...
        how to use (example):
        >>  uf = UnionFindTree(N) 
        >>  if uf.find_root(a) == uf.find_root(b):
        >>      do something
        >>  else:
        >>      do something
        >>      uf.unite(a, b)
        """
        def __init__(self, N):
            self.root = [-1] * (N+1)
            self.rank   = [0]  * (N+1)
            self.connected_num = [1] * (N+1)
    
        def find_root(self,x):
            root = self.root
            while root[x] != -1:
                x = root[x]
            return x
    
        def unite(self,x,y):
            root = self.root
            rank = self.rank
            connected_num = self.connected_num
            find_root = self.find_root
    
            rx = find_root(x)
            ry = find_root(y)
            if rx != ry:
                if rank[rx] < rank[ry]:
                    root[rx] = ry
                    rx,ry = ry,rx
                else:
                    if rank[rx] == rank[ry]:
                        rank[rx] += 1
                    root[ry] = rx       
                connected_num[rx] += connected_num[ry]

    class SegTree:
        """
        ref: https://qiita.com/takayg1/items/c811bd07c21923d7ec69
        init(init_val, ide_ele): 配列init_valで初期化 O(N)
        update(k, x): k番目の値をxに更新 O(logN)
        query(l, r): 区間[l, r)をsegfuncしたものを返す O(logN)
        """
        def __init__(self, init_val, segfunc, ide_ele):
            """
            init_val: 配列の初期値
            segfunc: 区間にしたい操作
            ide_ele: 単位元
            n: 要素数
            num: n以上の最小の2のべき乗
            tree: セグメント木(1-index)
            """
            n = len(init_val)
            self.segfunc = segfunc
            self.ide_ele = ide_ele
            self.num = 1 << (n - 1).bit_length()
            self.tree = [ide_ele] * 2 * self.num
            # 配列の値を葉にセット
            for i in range(n):
                self.tree[self.num + i] = init_val[i]
            # 構築していく
            for i in range(self.num - 1, 0, -1):
                self.tree[i] = self.segfunc(self.tree[2 * i], self.tree[2 * i + 1])

        def update(self, k, x):
            """
            k番目の値をxに更新
            k: index(0-index)
            x: update value
            """
            k += self.num
            self.tree[k] = x
            while k > 1:
                self.tree[k >> 1] = self.segfunc(self.tree[k], self.tree[k ^ 1])
                k >>= 1

        def query(self, l, r):
            """
            [l, r)のsegfuncしたものを得る
            l: index(0-index)
            r: index(0-index)
            """
            res = self.ide_ele
            l += self.num
            r += self.num
            while l < r:
                if l & 1:
                    res = self.segfunc(res, self.tree[l])
                    l += 1
                if r & 1:
                    res = self.segfunc(res, self.tree[r - 1])
                l >>= 1
                r >>= 1
            return res

    # ************ Graph ************
    # Graph: https://en.wikipedia.org/wiki/Directed_graph
    # 最短経路問題: https://ja.wikipedia.org/wiki/%E6%9C%80%E7%9F%AD%E7%B5%8C%E8%B7%AF%E5%95%8F%E9%A1%8C
    
    # Bellman-Ford: O(|V||E|). Use this if there exists an edge with negative length in the graph
    # After N steps, the shortest path has converded if there doesn't exist an cycle of edges with negative
    # Watch out: d[N] == d[2*N] doesn't necessarily mean the graph doesn't have negative cycle
    # ref: https://www.youtube.com/watch?v=1Z6ofKN03_Y

    def BellmanFord(N, M, ABC, vertex_start, vertex_end, value_if_inf = -1, find_shortest = False):
        """to calculate furthest or shortest length between vertex_start and vertex_end using BellmanFord algorithm
        
        Args:
            N (int): number of vertices
            M (int): number of edges
            ABC (list): [(ai, bi, ci) for _ in range(N)] where i-th edge is directed from vertex ai to vertex bi and the length is ci 
            vertex_start (int): start vertex. usually use 0.
            vertex_end (int): end vertex. usually use N-1.
            value_if_inf (int or string as you like, optional): value you want when the furthest (or shortest) distance is infinite (or -infinite). Defaults to -1.
            find_shortest (bool, optional): choose False to find furthest path. Defaults to False.
        
        Returns:
            int or string: normally int (but can be str if you set value_if_inf to str)
            
        Example:
                N, M, P = R()
                ABC = [R() for _ in range(M)]
                ABC = [(a-1, b-1, c-P) for a, b, c in ABC]
                print(BellmanFord(N, M, ABC, 0, N-1, value_if_inf = 'inf'))

        """

        def make_reachable_list(N, M, ABC, vertex_start, vertex_end):
            reachable_to_direct = defaultdict(list)
            reachable_from_direct = defaultdict(list)
            reachable_from_start = [False] * N
            reachable_to_end = [False] * N
            reachable_from_start[vertex_start] = True
            reachable_to_end[vertex_end] = True
            reachable_from_both_sides = [False] * N
            dfs_from_start = []
            dfs_to_end = []
            for a, b, c in ABC:
                reachable_to_direct[a].append(b)
                reachable_from_direct[b].append(a)
                if a == vertex_start:
                    dfs_from_start.append(b)
                    reachable_from_start[b] = True
                if b == vertex_end:
                    dfs_to_end.append(a)
                    reachable_to_end[a] = True
            while dfs_from_start:
                v = dfs_from_start.pop()
                for i in reachable_to_direct[v]:
                    if not reachable_from_start[i]:
                        reachable_from_start[i] = True
                        dfs_from_start.append(i)
            while dfs_to_end:
                v = dfs_to_end.pop()
                for i in reachable_from_direct[v]:
                    if not reachable_to_end[i]:
                        reachable_to_end[i] = True
                        dfs_to_end.append(i)
            for i in range(N):
                reachable_from_both_sides[i] = reachable_from_start[i] and reachable_to_end[i]
            return reachable_from_both_sides
        
        reachable_from_both_sides = make_reachable_list(N, M, ABC, vertex_start, vertex_end)
        
        if find_shortest:
            dist = [INF for i in range(N)]
        else:
            dist = [-INF for i in range(N)]
        
        dist[vertex_start] = 0
        for i in range(N):
            updated = False
            for a, b, c in ABC:
                if not reachable_from_both_sides[a]:
                    continue
                elif find_shortest:
                    update_condition = dist[a] + c < dist[b]
                else:
                    update_condition = dist[a] + c > dist[b]
                if dist[a] != INF and update_condition:
                    dist[b] = dist[a] + c
                    updated = True
                    if i == N-1:
                        return value_if_inf
            if not updated:
                break
        return dist[vertex_end]

    # Warshall Floyd O(V**3)
    def warshall_floyd(number_vertex, XYD, shift_one_for_vertex=False):
        # ref: https://juppy.hatenablog.com/entry/2018/11/01/%E8%9F%BB%E6%9C%AC_python_%E5%85%A8%E7%82%B9%E5%AF%BE%E6%9C%80%E7%9F%AD%E7%B5%8C%E8%B7%AF%E6%B3%95%EF%BC%88%E3%83%AF%E3%83%BC%E3%82%B7%E3%83%A3%E3%83%AB%E3%83%95%E3%83%AD%E3%82%A4%E3%83%89%E6%B3%95
        # number_vertex: len(vertex)
        # XYD: list of distance from x to y
        # shift_one_for_vertex: XYDの値が1~Nの時(0~N-1でない時)に調整するフラグ。
        # d[i][j]: iからjへの最短距離
        d = [[float("inf")]*number_vertex for i in range(number_vertex)]
        for x, y, dist in XYD:
            if shift_one_for_vertex:
                x = x - 1
                y = y - 1
            d[x][y] = dist
            d[y][x] = dist
        for i in range(number_vertex):
            d[i][i] = 0 # distance of same vertex is 0
        for k in range(number_vertex):
            for i in range(number_vertex):
                for j in range(number_vertex):
                    d[i][j] = min(d[i][j],d[i][k] + d[k][j])
        return d


    """ initialize variables and set inputs
    # initialize variables
        # to initialize list, use [0] * n
        # to initialize two dimentional array:
            # ex) [[0] * N for _ in range(N)]
            # ex2) dp = [[0] * (N+1) for _ in range(W*2)]
    # set inputs
        # put inputs between specific values (almost as quickly)
        # ex) S = [-INF] + [int(r()) for _ in range(A)] + [INF]
        # open(0).read() is sometimes useful:
        # ex) n, m, *x = map(int, open(0).read().split())
        #     min(x[::2]) - max(x[1::2])
        # ex2) *x, = map(int, open(0).read().split())
        #     don't forget to add comma after *x if only one variable is used
    # preprocessing
        # transpose = [x for x in zip(*data)]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
        # flat = [flatten for inner in data for flatten in inner]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [1, 2, 3, 4, 5, 6, 7, 8, 9]
    # calculate and output
        # output pattern
        # ex1) print(*l) => when l = [2, 5, 6], printed 2 5 6
    """

    # functions to read input
    r = lambda: sys.stdin.readline().strip()
    r_int = lambda: int(r())
    r_float = lambda: float(r())
    R = lambda: list(map(int, r().split()))
    R_str = lambda: r().split()
    R_map = lambda: map(int, r().split())
    R_float = lambda: list(map(float, r().split()))
    R_tuple = lambda: tuple(map(int, r().split()))

    """ how to treat input
    # single int: int(r())
    # single string: r()
    # single float: float(r())
    # line int: R()
    # line string: r().split()
    # line (str, int, int): [j if i == 0 else int(j) for i, j in enumerate(r().split())]
    # lines int: [R() for _ in range(n)]
    """

    # for test
    if sample_file:
        sys.stdin = open(sample_file)
        
    # ----------------------------------
    # main
    #%%
    N = r_int()
    AB = [R() for _ in range(N-1)]
    tree = [[] for _ in range(N)]
    for a, b in AB:
        tree[a-1].append(b-1)
        tree[b-1].append(a-1)
    task = [(i, [0]) for i in tree[0]]
    arrived = [0] * N
    arrived[0] = 1
    while task:
        node, parent = task.pop()
        for child in tree[node]:
            if child == N-1:
                way = parent + [node, N-1]
                break
            if arrived[child] == 0:
                arrived[child] = 1
                task.append((child, parent+[node]))
    kireme = way[(len(way)-1)//2]
    kireme_ikenai = way[(len(way)-1)//2+1]
    
    task = [0]
    arrived = [0] * N
    arrived[0] = 1
    cnt = 1
    while task:
        node = task.pop()
        for child in tree[node]:
            if child == kireme_ikenai:
                continue
            else:
                if not arrived[child]:
                    cnt += 1
                    arrived[child] = 1
                    task.append(child)
    if cnt > N - cnt:
        print('Fennec')
    else:
        print('Snuke')
        

        
    
    
    # 小文字をカウントしたリストを持つセグ木で行けそう。

    
    # a = [14, 5, 9, 13, 7, 12, 11, 1, 7, 8]
    


    #%%
    # end of main
    # ----------------------------------

    """memo: how to use defaultdict of list
    # initialize
    Dic = defaultdict(list)
    # append / extend
    Dic[x].append(y)
    # three methods for loop: keys(), values(), items()
    for k, v in Dic.items():
    """
    
    """memo: how to solve binary problems
    # to make binary digits text
    >>> a = 5
    >>> bin_str_a = format(a, '#06b')
    >>> print(bin_str_a)
    0b0101 # first 2 strings (='0b') indicates it is binary
    """
    
    """memo: how to solve the problem
    creating simple test/answer
    greed
    simple dp
    graph
    """

if __name__ == '__main__':
    main()