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;using ll=long long;using vi=vector<int>;using vvi=vector<vi>;using vl=vector<ll>;using vvl=vector<vl>;using P=pair<int,int>;using PL=pair<ll,ll>;using vp=vector<P>;using vpl=vector<PL>;
#define pqueue priority_queue
template<typename T>constexpr auto inf=numeric_limits<T>::max()/2;constexpr int INF=inf<int>,MOD=1000000007;constexpr ll LINF=inf<ll>;
#define _ol3(_1,_2,_3,name,...)name
#define _rep(i,n)_repi(i,0,n)
#define _repi(i,a,b)for(int i=a,i##_l=(b);i<i##_l;++i)
#define REP(...)_ol3(__VA_ARGS__,_repi,_rep,)(__VA_ARGS__)
#define REPR(i,n)for(int i=n-1;i>=0;--i)
#define REPA(i,v)REP(i,(v).size())
#define all(v)(v).begin(),(v).end()
#define rall(v)(v).rbegin(),(v).rend()
#define bit(n)(1ll<<(n))
#define F first
#define S second
#define endl '\n'
#define cho(n,a,b)cout<<((n)?a:b)<<endl
void YES(int n){cho(n,"YES","NO");}void Yes(int n){cho(n,"Yes","No");}void Poss(int n){cho(n,"Possible","Impossible");}
void _print(ostream&){}template<class T,class...U>void _print(ostream&s,const T&t,const U&...u){s<<t<<(sizeof...(u)?' ':'\n');_print(s,u...);}
template<class...T>void print(const T&...t){_print(cout,t...);}template<class...T>void dprint(const T&...t){_print(cerr,t...);}
#ifndef LOCAL
struct osd{template<class T>osd&operator<<(const T&t){return*this;}};osd cer_;
#define dprint(...)
#define cerr cer_
#endif
#define dbg(x) dprint("@l",__LINE__,':',#x,'=',x)
template<class T>bool chmax(T&a,const T&b){bool x=a<b;x?a=b:b;return x;}template<class T>bool chmin(T&a,const T&b){bool x=a>b;x?a=b:b;return x;}
template<class T,class U>ostream&operator<<(ostream&o,const pair<T,U>&p){return o<<p.F<<' '<<p.S;}
template<class T,class U>istream&operator>>(istream&i,pair<T,U>&p){return i>>p.F>>p.S;}
template<class T>class iterable{static false_type c(string v);template<class U>static auto c(U v)->decltype(all(v),true_type());static false_type c(...);public:const static bool value=decltype(c(declval<T>()))::value;};
template<class T,enable_if_t<iterable<T>::value,int> =0>ostream&operator<<(ostream&o,const T&v){for(auto&&i:v)o<<i<<' ';return o;}
template<class T>istream&operator>>(istream&i,vector<T>&v){for(T&j:v)i>>j;return i;}template<class T>vector<T>&operator<<(vector<T>&v,const T&t){v.push_back(t);return v;}
template<class T>queue<T>&operator<<(queue<T>&v, const T&t){v.push(t);return v;}template<class T>queue<T>&operator>>(queue<T>&v, T&t){t=v.front();v.pop();return v;}
template<class T>pqueue<T>&operator<<(pqueue<T>&v, const T&t){v.push(t);return v;}template<class T>pqueue<T>&operator>>(pqueue<T>&v, T&t){t=v.front();v.pop();return v;}
template<class T>stack<T>&operator<<(stack<T>&v, const T&t){v.push(t);return v;}template<class T>stack<T>&operator>>(stack<T>&v, T&t){t=v.front();v.pop();return v;}
template<class T>set<T>&operator<<(set<T>&v,const T&t){v.insert(t);return v;}template<class T>multiset<T>&operator<<(multiset<T>&v,const T&t){v.insert(t);return v;}
template<class T,enable_if_t<is_integral<T>::value,int> =0>T mod(T a, T b){if(a>0)return a%b;return b-(-a%b);}
template<class T,enable_if_t<is_floating_point<T>::value,int> =0>T mod(T a, T b){if(a>0)return a%b;return b-fmod(-a,b);}
template<class T,class U>pair<T,U>&operator+=(pair<T,U>&a, pair<T,U>&b){a.F+=b.F;a.S+=b.S;return a;}template<class T,class U>pair<T,U>operator+(const pair<T,U>&a, const pair<T,U>&b){T r=a;return r+=b;}
template<class T,class U>pair<T,U>&operator-=(pair<T,U>&a, pair<T,U>&b){a.F-=b.F;a.S-=b.S;return a;}template<class T,class U>pair<T,U>operator-(const pair<T,U>&a, const pair<T,U>&b){T r=a;return r-=b;}
template<class T,class U>pair<T,U>operator-(pair<T,U>&a){return mp(-a.F, -a.S);}

using len_type=ll;

vvi graph;
void dijkstra(int start, vector<len_type>& dist){
    int n=graph.size();
    fill(all(dist), inf<len_type>);
    dist.resize(n, inf<len_type>);
    priority_queue<P,vector<P>,greater<P>> q;
    dist[start]=0;
    q.push(make_pair(0,start));
    while(!q.empty()){
        P p=q.top();
        q.pop();
        int v=p.second;
        if(dist[v]<p.first)continue;
        for(const int&e:graph[v]){
            if(dist[e]>dist[v]+1){
                dist[e]=dist[v]+1;
                q.push(make_pair(dist[e],e));
            }
        }
    }
}
class UnionFind{
private:
    vi data;
public:
    UnionFind(int size):data(size,-1){}
    bool unite(int x,int y){
        x=root(x);y=root(y);
        if(x!=y){
            if(data[y]<data[x])swap(x,y);
            data[x]+=data[y];data[y]=x;
        }
        return x!=y;
    }
    bool find(int x,int y){
        return root(x)==root(y);
    }
    int root(int x){
        return data[x]<0?x:data[x]=root(data[x]);
    }
    int size(int x){
        return -data[root(x)];
    }
};
int main(){
    cin.tie(0);ios::sync_with_stdio(0);
    cout<<setprecision(10);
    int n,u,v;
    cin>>n>>u>>v;
    --u;--v;
    graph.resize(n);
    REP(i,n-1){
        int a,b;
        cin>>a>>b;
        --a;--b;
        graph[a]<<b;
        graph[b]<<a;
    }
    vl du,dv;
    dijkstra(u,du);
    dijkstra(v,dv);
    // bool f=(dv[u]%2);



    // REP(i,n)x[i]=(du[i]<dv[i]);
    vi d(n,-1);
    d[u]=0;
    queue<int> q;
    q<<u;
    while(q.size()){
        int ds;
        q>>ds;
        for(int e:graph[ds]){
            if(d[e]==-1){
                if(du[e]<=dv[e]){
                    d[e]=d[ds]+1;
                    if(du[e]<dv[e])
                        q<<e;
                }
            }
        }
    }
    //dbg(d);
    ll ma=0;
    REP(i,n)if(d[i]>=0)chmax(ma, dv[i]);
    print(ma-1);
    //cout<<*max_element(all(d));


}