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

Solution in Ruby (2.7.1) (judge verdict: Wrong Answer)

# 無向グラフ
class Graph
  attr_accessor *(?a..?z).to_a.map(&:to_sym)
  def initialize(n,pairs)
    @n = n
    @g = Array.new(n){ [] }
    pairs.each do |a,b|
      g[a] << b
      g[b] << a
    end
  end

  def [](v)
    g[v]
  end
end

# 有向グラフ
class DiGraph < Graph
  def initialize(n,pairs,root)
    super(n,pairs)
    @root = root
    @h = @g.dup
    @g = Array.new(n){ [] }
    seen = Array.new(n, false)
    que = [root]
    seen[root] = true
    while que.size > 0
      v = que.shift
      h[v].each do |nv|
        next if seen[nv]
        seen[nv] = true
        g[v] << nv
        que << nv
      end
    end
  end

  def dist
    ans = Array.new(n,nil)
    que = [[@root,0]]
    while que.size > 0
      v, depth = que.shift
      ans[v] = depth
      g[v].each do |nv|
        que << [nv, depth+1]
      end
    end
    ans
  end
end

class Problem
  attr_accessor *(?a..?z).to_a.map(&:to_sym)
  def initialize
    @n = gets.to_s.to_i
    @p = Array.new(n-1){ gets.split.map(&:to_i).map{ _1 - 1 } }
  end

  def solve
    fe = DiGraph.new(n,p,0)
    su = DiGraph.new(n,p,n-1)
    rank = fe.dist.zip(su.dist)
    cnt = rank.count{|i,j| i <= j}
    n <= cnt * 2
  end

  def show(ans)
    puts ans ? "Fennec" : "Snuke"
  end
end

Problem.new.instance_eval do
  show(solve)
end