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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.*;
import java.util.*;
import java.lang.*;
class Graph
{
	public ArrayList adjacency[];
	public Graph(int nodes)
	{
		adjacency = new ArrayList[nodes];
		for(int i=0;i<nodes;i++)
		{
			adjacency[i] = new ArrayList<Integer>();
		}
	}
	@Override
	public String toString()
	{
		String ret = "";
		for (ArrayList temp : adjacency) {
			ret+=(temp.toString()+"\n");
		}
		return ret;
	}
	public void addEdge(int from, int to)
	{
		adjacency[from].add(to);
		adjacency[to].add(from);
	}
}
	
public class Main {
	public static Graph g;
	public static int taka,aoki;
	public static int count[];
	public static boolean vis[];
	public static void mark(int node, int pos)
	{
		count[node] = pos;
		ArrayList list = g.adjacency[node];
		int val;
		for(int i=0;i<list.size();i++)
		{
			val = (int)list.get(i);
			if(count[val]<0)
			{
				mark(val,pos+1);
			}
		}
	}
	public static int calc(int node, int pos,PrintWriter out)
	{
		vis[node]=true;
		if( count[node]+1==pos)
			return count[node];
		// if(count[node]<pos)
		// 	return Integer.MIN_VALUE;
		ArrayList list = g.adjacency[node];
		// out.println("node: "+node);
		// out.println("list: "+list);
		// out.println();
		int val;
		int ans = 0;
		if(list.size()==1){
			// out.println("yo");
			ans = count[(int)list.get(0)];
		}
		for(int i=0;i<list.size();i++)
		{
			val = (int)list.get(i);
			if(!vis[val])
				ans=Math.max(ans,calc(val,pos+1,out));
		}
		return ans;
	}
	public static void main(String[] args)throws IOException {
        InputReader in = new InputReader(System.in);
        PrintWriter out = new PrintWriter(System.out);
        
		int n = in.nextInt();
		taka = in.nextInt()-1;
		aoki = in.nextInt()-1;
		g = new Graph(n);
		int a,b;
		for(int i=0;i<n-1;i++)
		{
			a=in.nextInt()-1;
			b=in.nextInt()-1;
			g.addEdge(a,b);
		}
		count = new int[n];
		vis = new boolean[n];
		// Arrays.fill(count,-1);
		for(int i=0;i<n;i++)
		{
			count[i]=-1;
			vis[i]=false;
		}
		mark(aoki,0);
		// out.println("marked");
		// for(int i=0;i<n;i++)
		// {
		// 	out.print(i+" ");
		// 	out.println(count[i]);
		// }
		out.println(calc(taka,0,out));
        out.close();
	}
    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

    }

}