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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i,N) for(int i=0;i<N;i++)
#define rep2(i,N) for(int i=1;i<=N;i++)
using namespace std;
long long  INF=1000000000000000000;
long long mod=1000000007;



int N,u,v;
vector<vector<int>>edge(100010);
int distance_u[100010];
int distance_v[100010];


void dist_u(int u){

    queue<int>q;

    q.push(u);

    bool reached[100010]={};
    
    rep(i,N)distance_u[i]=1e9;

    reached[u]=true;
    distance_u[u]=0;
    while(!q.empty()){

        int now=q.front();
        q.pop();
        
        if(now==v)continue;

        for(auto nv:edge[now]){
            if(reached[nv])continue;
            distance_u[nv]=distance_u[now]+1;
            reached[nv]=true;
            q.push(nv);
        }
    }

    return ;
}

void dist_v(int v){
    queue<int>q;

    q.push(v);

    bool reached[100010]={};

    rep(i,N)distance_v[i]=1e9;

    distance_v[v]=0;
    reached[v]=true;
    while(!q.empty()){
        int now=q.front();
        q.pop();
        
        

        for(auto nv:edge[now]){
            if(reached[nv])continue;
            distance_v[nv]=distance_v[now]+1;
            reached[nv]=true;
            q.push(nv);
        }
    }

    return;

}


int main(){


cin>>N>>u>>v;
u--;
v--;

distance_v[v]=0;



rep(i,N-1){
    int a,b;
    cin>>a>>b;
    a--,b--;
    edge[a].emplace_back(b);
    edge[b].emplace_back(a);
}

if(edge[u].size()==1 && edge[u][0]==1){
    cout<<"0"<<endl;
    return 0;
}


/*
(i)最初の行動でTがAのいる頂点にしか移動できない時=>ans=0

(ⅱ)そうでない時
TとAはすれ違えないので、ゲームが終了するのはTが行き止まりに追い詰められた時。
言い換えると、Aの初期位置を根として、Tが葉xにいてAがxの親の親である時、
それぞれ一回の移動でゲームが終了する。

Aは最短距離でTを追跡できるのでAの移動回数はTが追い詰められる葉xの位置によって決まる。

ここで葉xについて、葉xはTがAよりも早く到達できる葉の中で、
Tの初期位置からもっとも遠い葉になる。

そのような葉xをDFSまたはBFSで求める。

*/
vector<int>cand;

rep(i,N){
    if(edge[i].size()==1){
        cand.emplace_back(i);
    }
}

int ans=-1;

dist_u(u);
dist_v(v);
/*
rep(i,N)cout<<distance_u[i]<<" ";
cout<<endl;



rep(i,N)cout<<distance_v[i]<<" ";
cout<<endl;
*/


for(auto nv:cand){
    //cout<<(distance_u[nv])<<" "<<distance_v[nv]<<endl;
    if(distance_u[nv]<distance_v[nv]){
        ans=max(ans,distance_v[nv]-1);
    }
}

cout<<ans<<endl;



return 0;

}
/*
5 4 1
1 2
2 3
3 4
3 5

2

5 4 5
1 2
1 3
1 4
1 5

1

2 1 2
1 2

0

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

5
*/