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 PyPy3 (2.4.0) (judge verdict: Accepted)

#メモ，葉の一つ手前でぶつかるはず？
#uの位置を無視していたけど，例えばuが葉，vがその一つ手前にいる場合に他の葉の深さが関係なくなる，やはり根付きとかにする？
import heapq
class Dijkstra():
    """
    ・有向 / 無向は問わない(無向の場合は，逆向きの辺もたす)
    ・負のコストがない場合のみ
    ・計算量はO(E log|V|)　
    ・heapを使うことで頂点を走査する必要がなくなる（代わりに，距離更新したものは確定でなくともqueに入れておく）
    """
    
    #最短のpathをたす
    
    
    class Edge():
        #重み付き有向辺
        
        def __init__(self, _to, _cost):
            self.to =_to
            self.cost = _cost
    
    def __init__(self, V):
        #引数Vは頂点数
        self.G = [[] for _ in range(V)] #隣接リストG[u][i]が頂点uのi番目の辺
        self. _E = 0 #辺の数
        self._V = V #頂点数
    
    #proparty - 辺の数
    def E(self):
        return self._E
    
    #proparty - 頂点数
    def V(self):
        return self._V
    
    def add(self, _from, _to, _cost):
        #2頂点と辺のコストを追加
        self.G[_from].append(self.Edge(_to,_cost))
        self._E +=1
        
    def add2(self, _from, _to, _cost):
        #2頂点と辺のコスト（無向）を追加
        self.G[_from].append(self.Edge(_to, _cost))
        self.G[_to].append(self.Edge(_from, _cost))
        self._E +=1
        
    def shortest_path(self,s):
        #始点sから頂点iまでの最短経路長のリストを返す
        
        que = [] #priority queue
        d = [float("inf")] * self.V()
        prev = [None]*self.V() #prev[j]は，sからjへ最短経路で行くときのjの一つ前の場所
        d[s] = 0
        heapq.heappush(que,(0,s)) #始点の距離と頂点番号をヒープに追加
        
        while len(que)!=0:
            #キューに格納されてある中で一番コストが小さい頂点を取り出す
            cost,v = heapq.heappop(que)
            
            #キューに格納された最短経路長候補がdの距離よりも大きい場合に処理をスキップ
            if d[v] < cost:
                continue
                
            #頂点vに隣接する各頂点iに対して，vを経由した場合の距離を計算して，これがd[i]よりも小さい場合に更新
            for i in range(len(self.G[v])):
                e = self.G[v][i] #vのi個目の隣接辺
                if d[e.to] > d[v] + e.cost:
                    d[e.to] = d[v] + e.cost #更新
                    prev[e.to] = v
                    heapq.heappush(que,(d[e.to],e.to)) #queに新たな最短経路長候補を追加
                    
        return d
#ここまでダイクストラ


N, u, v = map(int, input().split())
list=[[] for _ in range(N)]
djk = Dijkstra(N)

#a,bなどは1からスタートするので1引いておく
for i in range(N-1):
    a,b  = map(int, input().split())
    list[a-1].append(b-1)
    list[b-1].append(a-1)
    djk.add2(a-1,b-1,1)


#青木くんの位置(v-1)から各葉までの距離を出す
d_v=djk.shortest_path(v-1)
d_u=djk.shortest_path(u-1)

#青木くんより高橋くんが先に到達できる葉でなければならない？
path_to_leaf_v=[]

for i in range(N):
    if len(list[i])==1:
        if d_v[i] >= d_u[i]:
            path_to_leaf_v.append(d_v[i])


#上記の中で，一番深い葉に高橋くんが逃げ込む
ans = max(path_to_leaf_v)-1
print(ans)