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 ff first
#define ss second
#define pb push_back
#define pf push_front
#define eb emplace_back
#define emp emplace
#define ins insert
#define mp make_pair
#define mt make_tuple
#define sz(s) (int)s.size()
#define forp(i,a,b) for( int i=a;i<=b;i++)
#define rep(i,n)    for( int i=0;i<n;i++)
#define ren(i,n)    for( int i=n-1;i>=0;i--)
#define forn(i,a,b) for( int i=a;i>=b;i--)
#define w(t) while(t)
#define sc(a) scanf("%d", &a);
#define on cout<<"\n"
#define os cout <<" "
#define o2(a,b) cout<<a<<" "<<b
#define o(a) cout << a
#define ppr(a) cout << a.first << " " << a.second
#define bitcount __builtin_popcount
#define gcd __gcd
#define all(v) v.begin(),v.end()
#define mem(n,m) memset(n,m,sizeof(n))
#define pii pair<int,int>
#define tiii tuple<int, int, int>
#define pll pair<long long,long long>
#define sii set<int>
#define sll set<long long>
#define vii vector<int>
#define vll vector<long long>
#define mii map<int,int>
#define mll map<long long, long long>
#define lob lower_bound
#define upb upper_bound
#define ret return
#define present(s,x) (s.find(x) != s.end())
#define cpresent(s,x) (find(all(s),x) != s.end())
#define ford(container, it) for(auto it = container.begin(); it != container.end(); it++)
#define fors(container, it, a, b) for(auto it = a; it != b; it++)
#define boost ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define MOD 1000000007
#define EPSILON 1e-9
#define PI acos(-1)
#define inf 1e9
#define int long long
#define debug(a) cerr<<#a<<": ";for(auto i:a)cerr<<i<<" ";cerr<<'\n';
#define trace(a) cerr<<#a<<": "<<a<<"\n"
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>


typedef long long  ll;
typedef long double  ldo;
typedef double  db ;
using namespace std;
auto start = std::chrono::system_clock::now();
inline void skj()
{
  std::chrono::time_point<std::chrono::system_clock> end;
  end = std::chrono::system_clock::now();
  std::chrono::duration<double> elapsed_seconds = end - start;
  std::time_t end_time = std::chrono::system_clock::to_time_t(end);
  cerr<<"Time taken " << elapsed_seconds.count()*1000<<"\n";
}

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
//uniform_int_distribution<int> ud(1, 100); use this for random number generator

    //custom hash for unordered map
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);
    }
};
inline int binexp(int a,int b,int m)
{
    if(a == 0) {
        ret 0;
    }
    int res=1;
    a%=m;
    while(b)
    {
        if(b&1)
            res=(res*1ll*a)%m;
        a=(a*1ll*a)%m;
        b>>=1;
    }
    return res;
}
const int N = 1e5+5;
int level[N], mxl[N], pr[N];
vii g[N];
inline void dfs(int s, int p) {
    mxl[s] = level[s];
    pr[s] = p;
    for(int i : g[s]) {
        if(i != p) {
            level[i] = level[s] + 1;
            dfs(i, s);
            mxl[s] = max(mxl[s], mxl[i]);
        }
    }
}
inline void solve_me_senpai() {
    int n, u, v;
    cin >> n >> u >> v;
    rep(i, n - 1) {
        int a, b;
        cin >> a >> b;
        g[a].pb(b);
        g[b].pb(a);
    }
    dfs(v, -1);
    int d = (level[u] - level[v] - 1) / 2;
    if((level[u] - level[v] - 1) % 2 && d) {
    //    d--;
    }
    //cout << d << " " << u << "\n";
    w(d--) {
        u = pr[u];
    }
    w(1) {
        int mx = -1;
        for(int x : g[u]) {
            if(x != pr[u]) {
                if(mx == -1 || (mxl[mx] < mxl[x])) {
                    mx = x;
                }
            }
        }
        if(mx == -1) {
            break;
        }
        u = mx;
    }
   // o(v);on;
    o(level[u] - 1);
}
signed main()
{
    boost
    #ifndef ONLINE_JUDGE
   // freopen("input.txt", "r", stdin);
  //  freopen("output.txt", "w", stdout);
    #endif
    int t = 1;
    //cin >> t;
    int a = 1;
    while(t--) {
        //cout << "Case #" << a  << ": ";
        solve_me_senpai();
        a++;
    }
    //skj();
    return 0;
}