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>
using namespace std;

#define ipr 			pair<int, int>
#define pb 				push_back
#define ff 				first
#define ss 				second
#define mp 				make_pair
#define fr(i, j, k) 	for (int i = j; i < k; i++)
#define rf(i, j, k) 	for (int i = j; i >= k; i--)
#define int 			long long
#define ll 				unsigned long long
#define ld 				long double
#define all(x) 			x.begin(), x.end()
#define rall(x) 		x.rbegin(), x.rend()
#define cnt_ones(x) 	__builtin_popcountll(x)
#define dbg() 			cerr << "okay" << endl;			

#define IOS                           \
	std::ios::sync_with_stdio(false); \
	cin.tie(NULL);                    \
	cout.tie(NULL);

const int inf = 1000000000000000000;
const int Maxn = 100001;
const ld pi = acos(-1);
const int mod = 998244353;

template <class T>
ostream &operator<<(ostream &out, vector<T> &A)
{
	for (auto x : A)
		out << x << " ";
	return out;
}

int n, e;
vector<vector<int > > g;
vector<vector<int>> dis;

void graph_print(){
    fr(i,0,n)
        cout << g[i] << endl;
}

void graph_inp(){
    g.resize(n);
    e = n-1;    //for a tree
    fr(i,0,e){
        int u,v;    cin>>u>>v;
        --u;   --v;
        g[u].pb(v);
        g[v].pb(u);
    }  
    
}

void bfs(int root, int col){
    queue<int> q;   q.push(root);
    dis[root][col] = 0;
    while(!q.empty()){
        int node = q.front();   q.pop();
        for(auto child : g[node])
            if(dis[child][col] == -1){
                dis[child][col] = dis[node][col] + 1;
                q.push(child);
            }
          
    }
}

int32_t main(){IOS
    int x,y, ans=0;
    cin>>n>>x>>y;  x--; y--;
    graph_inp();
    dis.resize(n, vector<int>(2, -1));
    bfs(y,0);   bfs(x,1);
    fr(i,0,n)
       if(dis[i][1] < dis[i][0])
            ans = max(ans, dis[i][0]);
    cout << ans-1 << endl;
	return 0;
}