Problem:
Score : 600 points

Problem Statement

We have a tree with N vertices. The i-th edge connects Vertex A_i and B_i bidirectionally.

Takahashi is standing at Vertex u, and Aoki is standing at Vertex v.

Now, they will play a game of tag as follows:

1. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Takahashi moves to a vertex of his choice that is adjacent to his current vertex.

2. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Aoki moves to a vertex of his choice that is adjacent to his current vertex.

3. Go back to step 1.

Takahashi performs his moves so that the game ends as late as possible, while Aoki performs his moves so that the game ends as early as possible.

Find the number of moves Aoki will perform before the end of the game if both Takahashi and Aoki know each other's position and strategy.

It can be proved that the game is bound to end.

Constraints

2 \leq N \leq 10^5

1 \leq u,v \leq N

u \neq v

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 u v
A_1 B_1
:
A_{N-1} B_{N-1}

Output

Print the number of moves Aoki will perform before the end of the game.

Sample Input 1

5 4 1
1 2
2 3
3 4
3 5

Sample Output 1

2

If both players play optimally, the game will progress as follows:

Takahashi moves to Vertex 3.

Aoki moves to Vertex 2.

Takahashi moves to Vertex 5.

Aoki moves to Vertex 3.

Takahashi moves to Vertex 3.

Here, Aoki performs two moves.

Note that, in each move, it is prohibited to stay at the current vertex.

Sample Input 2

5 4 5
1 2
1 3
1 4
1 5

Sample Output 2

1

Sample Input 3

2 1 2
1 2

Sample Output 3

0

Sample Input 4

9 6 1
1 2
2 3
3 4
4 5
5 6
4 7
7 8
8 9

Sample Output 4

5

Solution in Ruby (2.3.3) (judge verdict: Wrong Answer)

de = false
require 'set'
n,u,v = gets.chomp.split.map(&:to_i)
edges = {}
(n-1).times do |i|
  a, b= gets.chomp.split.map(&:to_i)
  edges[a] ||= Set.new
  edges[a].add b
  edges[b] ||= Set.new
  edges[b].add a
end
p edges if de

taka = [Set.new([u])]
aoki = [Set.new([v])]
taka_visited = Array.new(n+1,-1)
aoki_visited = Array.new(n+1,-1)
taka_visited[u] = 0
aoki_visited[v] = 0
turn = 0

until aoki.empty?
  p turn if de
  p ['taka', taka_visited] if de
  p ['aoki', aoki_visited] if de
  turn += 1
  taka_v = taka.shift
  taka_next = Set.new
  break_flg = false

  taka_v.each do |from|
    if edges[from].size == 1 # 行き先が一個の場合強制ターンする
      to = edges[from].first

      if aoki_visited[to] >= 0
        break_flg = true
        turn -= 1
        break
      end
      taka_visited[to] = turn.odd? ? -2 : -3
    else
      edges[from].each do |to|
        next if taka_visited[to] >= 0
        next if aoki_visited[to] >= 0
        taka_next.add to
        taka_visited[to] = turn
      end
    end
  end
  break if break_flg
  taka.push taka_next

  # -- Aoki
  aoki_v = aoki.shift
  aoki_next = Set.new
  aoki_v.each do |from|
    edges[from].each do |to|
      if taka_visited[to] == turn
        break_flg = true
        break
      elsif taka_visited[to] <= -2
        turn += 1 if (turn + taka_visited[to]).even?
        break_flg = true
        break
      end
      next if aoki_visited[to] >= 0
      aoki_next.add to
      aoki_visited[to] = turn
    end
  end
  break if break_flg
  aoki.push aoki_next
end

puts turn