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 LuaJIT (2.0.4) (judge verdict: Accepted)

local mfl = math.floor
local n = io.read("*n")
local edge = {}
for i = 1, n do edge[i] = {} end
for i = 1, n - 1 do
  local a, b = io.read("*n", "*n")
  table.insert(edge[a], b)
  table.insert(edge[b], a)
end

local tasks = {1}
local asked = {false}
local len = {0}
for i = 2, n do
  asked[i] = false
  len[i] = 0
end
local done = false
while not done do
  local src = tasks[#tasks]
  if asked[src] then
    table.remove(tasks)
  else
    asked[src] = true
    for i = 1, #edge[src] do
      local dst = edge[src][i]
      if not asked[dst] then
        len[dst] = len[src] + 1
        table.insert(tasks, dst)
        if dst == n then
          done = true
          break
        end
      end
    end
  end
end
local halflen = mfl(len[n] / 2)
local pos1, pos2 = 1, n
for i = 1, #tasks do
  local pos = tasks[i]
  local l = len[pos]
  if l == halflen then pos1 = pos
  elseif l == halflen + 1 then pos2 = pos end
end
tasks = {pos1, pos2}
for i = 1, n do asked[i] = 0 end
asked[pos1], asked[pos2] = 1, 2
while 0 ~= #tasks do
  local src = tasks[#tasks]
  table.remove(tasks)
  for i = 1, #edge[src] do
    local dst = edge[src][i]
    if 0 == asked[dst] then
      asked[dst] = asked[src]
      table.insert(tasks, dst)
    end
  end
end
local cnt = {0, 0}
for i = 1, n do
  cnt[asked[i]] = cnt[asked[i]] + 1
end
print(cnt[1] > cnt[2] and "Fennec" or "Snuke")