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,a,b) for (i=a;i<b;i++)
#define rep2(i,a,b) for (i=a;i>=b;i--)
#define mod 1000000007
#define mod2 998244353
#define FIO ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define INF  1e9+5
#define f first
#define s second
#define endl '\n'
#define ll long long
#define ii pair <int,int>
#define pll pair <ll,ll>
#define vi vector <int>
#define vl vector <ll>
#define vvi vector < vi >
#define vii vector < ii >
#define vpl vector < pll >
#define vvii vector < vii >
#define vll vector < ll >
#define vb  vector <bool>
#define pb push_back
#define mk make_pair
#define eb emplace_back
#define pg priority_queue<ll, vector<ll>, greater<ll>>
#define pq priority_queue<ll>
#define sz(a) a.size()
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define cset(a) __builtin_popcountll(a)
#define FILL(a, b, T) fill_n((T *)a, sizeof(a) / sizeof(T), b)
#define debug(x) cerr << #x << " is " << x << endl;
using namespace std;
#define error(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }
 
void err(istream_iterator<string> it) {}
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
    cerr << *it << " = " << a << endl;
    err(++it, args...);
}
struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
        // http://xorshift.di.unimi.it/splitmix64.c
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }
 
    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};
// unordered_map<long long, int, custom_hash> safe_map;
// FOR VIEWERS, CODE STARTS HERE
ll mex = 100005;
map<ll,vll>e;
vll d1(mex, INF), lg(mex, 0), lvl(mex);
ll ans1; 
void dfs1(ll u, ll par)
{
    ll tmp, val = INF;
    lvl[u] = lvl[par] + 1;
    for(auto p: e[u])
    {
        if( p == par)
            continue;
        dfs1(p, u);
        d1[u] = min(1 + d1[p], d1[u]);
        lg[u] = max(1 + lg[p], lg[u]);
    }
}
int main()
{
    // #ifndef ONLINE_JUDGE
    //  freopen("input.txt","r",stdin);
    //  freopen("output.txt","w",stdout);
    // #endif
    FIO;
    ll t, q, n, m, prev, maxx,  i, j, k1, k2, k3, k, x, y, flag, sum, cnt, ans;
    cin>>n>>k1>>k2;
    rep(i,0,n-1)
    {
        cin>>x>>y;
        e[x].pb(y);
        e[y].pb(x);
    }
    d1[k1] = 0;
    lvl[0] = -1;
    dfs1(k2,0);
    ans = 0;
    rep(i,1,n+1)
    {
        if( d1[i] < lvl[i])
        {
            x = lvl[i] + lg[i];
            // y = d1[i] + lg[i];
            // error(i, lvl[i], d1[i], lg[i]);
            ans = max(ans, x - 1);
            
        }
    }
    cout<<ans<<endl;
    return 0;    
}