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)

#!/usr/bin/env python3

import sys
import math
from bisect import bisect_right as br
from bisect import bisect_left as bl
sys.setrecursionlimit(1000000)
from heapq import heappush, heappop,heappushpop
from collections import defaultdict
from itertools import accumulate
from collections import Counter
from collections import deque
from operator import itemgetter
from itertools import permutations
mod = 10**9 + 7
inf = float('inf')
def I(): return int(sys.stdin.readline())
def LI(): return list(map(int,sys.stdin.readline().split()))

"""
[考察]
　高橋君が最も遠くに逃げる場合
　:= (高橋君が先に行ける頂点)Λ(青木君から最も遠い頂点)

⇒高橋君が先に行ける頂点
　高橋君からの各頂点の距離(高橋君を根としたときの木の深さ),
　青木君からの各頂点の距離(青木君を根としたときの木の深さ)を求めて、
　高橋君からの距離と青木君からの距離を比較
⇒青木君から最も遠い頂点
　青木君からの各頂点の距離の最大値
"""
n,V,U = LI()
V -= 1
U -= 1

edges = [[] for _ in range(n)]

for _ in range(n-1):
    a,b = LI()
    edges[a-1].append(b-1)
    edges[b-1].append(a-1)

if n == 2:
    print(0)
else:
    depth_v = [None]*n
    depth_v[V] = 0
    que = deque()
    que.append(V)
    while que:
        v = que.popleft()
        for u in edges[v]:
            if depth_v[u] == None:
                depth_v[u] = depth_v[v]+1
                que.append(u)
    
    depth_u = [None]*n
    depth_u[U] = 0
    que = deque()
    que.append(U)
    while que:
        u = que.popleft()
        for v in edges[u]:
            if depth_u[v] == None:
                depth_u[v] = depth_u[u] + 1
                que.append(v)
    
    ans = 0
    for v in range(n):
        if depth_v[v] < depth_u[v]:
            ans = max(ans,depth_u[v]-1)
    print(ans)