### BUGGY CODE ###

class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n

    def root(self, x):
        if self.parents[x] < 0:
            return x
        else:
            self.parents[x] = self.root(self.parents[x])
            return self.parents[x]

    def union(self, x, y):
        x = self.root(x)
        y = self.root(y)

        if x == y:
            return

        if self.parents[x] > self.parents[y]:
            x, y = y, x

        self.parents[x] += self.parents[y]
        self.parents[y] = x

    def size(self, x):
        return -self.parents[self.root(x)]

    def is_same(self, x, y):
        return self.root(x) == self.root(y)

    def members(self, x):  # 計算量n
        root = self.root(x)
        return [i for i in range(self.n) if self.root(i) == root]

    def all_roots(self):
        return [i for i, x in enumerate(self.parents) if x < 0]

    def group_count(self):
        return len(self.all_roots())

    def all_group_members(self):
        return {r: self.members(r) for r in self.all_roots()}

    def __str__(self):
        return ''.join('{}: {}'.format(r, self.members(r)) for r in self.all_roots())

n = int(input())
uf = UnionFind(n)

ab = [tuple(map(int,input().split())) for i in range(n-1)]
cost = [[float("inf") for i in range(n)] for i in range(n)]
for i, (a,b) in enumerate(ab):
    a,b = a-1,b-1
    cost[a][b]=1
    cost[b][a]=1

# cook your dish here
def dijkstra_back(s,t,n,w,cost):
    #sからtへの最短経路の経路復元
    prev = [s] * n #最短経路の直前の頂点
    d = [float("inf")] * n
    used = [False] * n
    d[s] = 0

    while True:
        v = -1
        for i in range(n):
            if (not used[i]) and (v == -1):
               v = i
            elif (not used[i]) and d[i] < d[v]:
                v = i
        if v == -1:
               break
        used[v] = True

        for i in range(n):
            if d[i] > d[v] + cost[v][i]:
                d[i] = d[v] + cost[v][i]
                prev[i] = v

    path = [t]
    while prev[t] != s:
        path.append(prev[t])
        prev[t] = prev[prev[t]]
    path.append(s)
    path = path[::-1]
    return path

route = dijkstra_back(0,n-1,n,n-1,cost)

mid = route[-(-len(route)//2)-1]
mid_next = route[-(-len(route)//2)]


for i, (a,b) in enumerate(ab):
    a,b = a-1,b-1
    if a == mid and b == mid_next:continue
    if a == mid_next and b == mid:continue

    uf.union(a,b)

if uf.size(0) > uf.size(n-1):
    print('Fennec')
else:
    print('Snuke')


### FIXED CODE ###

n = int(input())

ab = [tuple(map(int,input().split())) for i in range(n-1)]
es = [[] for i in range(n)]
for a,b in ab:
    a,b = a-1,b-1
    es[a].append(b)
    es[b].append(a)
INF = float('inf')
dist0 = [INF] * n
dist0[0] = 0
import heapq
hq = [(0,0)]
heapq.heapify(hq)
while hq:
    c,v = heapq.heappop(hq)
    for to in es[v]:
        if c+1 >= dist0[to]: continue
        dist0[to] = c+1
        heapq.heappush(hq,(c+1,to))

distn = [INF] * n
distn[n-1] = 0
hq = [(0,n-1)]
heapq.heapify(hq)
while hq:
    c,v = heapq.heappop(hq)
    for to in es[v]:
        if c+1 >= distn[to]: continue
        distn[to] = c+1
        heapq.heappush(hq,(c+1,to))


count_f = 0
for i in range(n):
    if dist0[i] <= distn[i]:
        count_f+=1

if count_f > n-count_f: print('Fennec')
else: print('Snuke')