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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;
using static Asakatsu20200317E.Input;
using static System.Math;

namespace Asakatsu20200317E
{
    class Input
    {
        private static String String() => Console.ReadLine();

        private static IEnumerable<string> Split() => String().Split(' ');

        private static long[] Long() => Split().Select(long.Parse).ToArray();

        /// <summary>
        /// 1行の文字列の入力
        /// </summary>
        public static void @in(out string s) => s = String();

        /// <summary>
        /// 複数行の文字列の入力
        /// </summary>
        public static void @in(long rowNumber, out string[] s)
        {
            s = new String[rowNumber];
            for (long i = 0; i < rowNumber; i++) @in(out s[i]);
        }

        /// <summary>
        /// 1行に書かれた1つの整数の入力
        /// </summary>
        public static void @in(out int i)
        {
            i = int.Parse(String());
        }

        /// <summary>
        /// 1行に書かれた1つの整数の入力
        /// </summary>
        public static void @in(out long a) => a = Long()[0];

        /// <summary>
        /// 1行に書かれた2つの整数の入力
        /// </summary>
        public static void @in(out long a, out long b)
        {
            long[] lArr = Long();
            a = lArr[0];
            b = lArr[1];
        }

        /// <summary>
        /// 1行に書かれた3つの整数の入力
        /// </summary>
        public static void @in(out long a, out long b, out long c)
        {
            long[] lArr = Long();
            a = lArr[0];
            b = lArr[1];
            c = lArr[2];
        }

        /// <summary>
        /// 1行に書かれた4つの整数の入力
        /// </summary>
        public static void @in(out long a, out long b, out long c, out long d)
        {
            long[] lArr = Long();
            a = lArr[0];
            b = lArr[1];
            c = lArr[2];
            d = lArr[3];
        }

        /// <summary>
        /// 1行に書かれた5つの整数の入力
        /// </summary>
        public static void @in(out long a, out long b, out long c, out long d, out long e)
        {
            long[] lArr = Long();
            a = lArr[0];
            b = lArr[1];
            c = lArr[2];
            d = lArr[3];
            e = lArr[4];
        }

        /// <summary>
        /// 1行に書かれた複数の整数列の入力
        /// </summary>
        public static void @in(out long[] lArr) => lArr = Long();

        /// <summary>
        /// rowNumber行に書かれた1つの整数列の入力
        /// </summary>
        public static void @in(long rowNumber, out long[] lArr)
        {
            lArr = new long[rowNumber];
            for (long i = 0; i < rowNumber; i++) @in(out lArr[i]);
        }

        /// <summary>
        /// rowNumber行に書かれた2つの整数列の入力
        /// </summary>
        public static void @in(long rowNumber, out long[] a, out long[] b)
        {
            a = new long[rowNumber];
            b = new long[rowNumber];
            for (int i = 0; i < rowNumber; i++) @in(out a[i], out b[i]);
        }

        /// <summary>
        /// rowNumber行に書かれた3つの整数列の入力
        /// </summary>
        public static void @in(long rowNumber, out long[] a, out long[] b, out long[] c)
        {
            a = new long[rowNumber];
            b = new long[rowNumber];
            c = new long[rowNumber];
            for (int i = 0; i < rowNumber; i++) @in(out a[i], out b[i], out c[i]);
        }

        /// <summary>
        /// rowNumber行に書かれた4つの整数列の入力
        /// </summary>
        public static void @in(long rowNumber, out long[] a, out long[] b, out long[] c, out long[] d)
        {
            a = new long[rowNumber];
            b = new long[rowNumber];
            c = new long[rowNumber];
            d = new long[rowNumber];
            for (int i = 0; i < rowNumber; i++) @in(out a[i], out b[i], out c[i], out d[i]);
        }

        /// <summary>
        /// h行w列の整数の行列の入力
        /// </summary>
        public static void @in(long h, long w, out long[][] a)
        {
            a = new long[h][];
            for (long i = 0; i < h; i++) @in(out a[i]);
        }

        /// <summary>
        /// 1行に書かれた1つの小数の入力
        /// </summary>
        public static void @in(out double d) => d = double.Parse(String());

        /// <summary>
        /// 1行に書かれた小数の配列の入力
        /// </summary>
        public static void @in(out double[] dArr) => dArr = Split().Select(double.Parse).ToArray();
    }

    class Program
    {
        public static void Main(string[] args)
        {
            StreamWriter streamWriter = new StreamWriter(Console.OpenStandardOutput()) {AutoFlush = false};
            Console.SetOut(streamWriter);
            new Solver().Solve();
            Console.Out.Flush();
        }

        public static void Debug()
        {
            new Solver().Solve();
        }
    }

    class Solver
    {
        private long N;
        private long u;
        private long v;
        private long[] A;
        private long[] B;

        public void Solve()
        {
            @in(out N, out u, out v);
            @in(N - 1, out A, out B);

            HashSet<long>[] tree = new HashSet<long>[N + 1];
            for (int i = 0; i < N + 1; i++)
            {
                tree[i] = new HashSet<long>();
            }

            for (int i = 0; i < N - 1; i++)
            {
                tree[A[i]].Add(B[i]);
                tree[B[i]].Add(A[i]);
            }


            long[] tDiff = new long[N + 1];
            long[] aDiff = new long[N + 1];

            Queue<long> q = new Queue<long>();
            HashSet<long> memo = new HashSet<long>();
            tDiff[u] = 0;
            memo.Add(u);
            q.Enqueue(u);

            while (q.Count > 0)
            {
                var c = q.Dequeue();
                foreach (long l in tree[c])
                {
                    if (memo.Contains(l))
                    {
                        continue;
                    }

                    memo.Add(l);
                    q.Enqueue(l);
                    tDiff[l] = tDiff[c] + 1;
                }
            }

            q = new Queue<long>();
            memo = new HashSet<long>();
            aDiff[v] = 0;
            memo.Add(v);
            q.Enqueue(v);

            while (q.Count > 0)
            {
                var c = q.Dequeue();
                foreach (long l in tree[c])
                {
                    if (memo.Contains(l))
                    {
                        continue;
                    }

                    memo.Add(l);
                    q.Enqueue(l);
                    aDiff[l] = aDiff[c] + 1;
                }
            }

            long ans = 0;
            for (int i = 1; i <= N; i++)
            {
                if (aDiff[i] > tDiff[i])
                {
                    ans = aDiff[i];
                }
            }

            Console.WriteLine(ans - 1);
        }
    }
}