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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

typedef long long int lli;
typedef unsigned long long int ulli;
typedef pair<int,int> pii;
typedef priority_queue< int, vector< int >, greater< int > >  minHeap;

#define ones(x) __builtin_popcount(x)
#define onesl(x) __builtin_popcountl(x)
#define onesll(x) __builtin_popcountll(x)

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define mt make_tuple
#define scn(n) scanf("%d",&n)
#define scnll(n) scanf("%lld",&n)
#define scn2(n,m) scanf("%d%d",&n,&m)
#define scn3(n,m,w) scanf("%d%d%d",&n,&m,&w)
#define scn2ll(n,m) scanf("%lld%lld",&n,&m)
#define atoz(v) v.begin(),v.end()
#define ratoz(v) v.rbegin(),v.rend()
#define Fill(a,v) memset(a,v,sizeof(a))
#define fi first
#define se second
#define inf 1e9
#define pi acos(-1.0)
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define mod 1000000007
#define FastIO ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define bug(x) cerr << __LINE__ << " says: " << #x << " = " << (x) << "\n"

int Set(int N,int pos)
{
    return N=N | (1<<pos);
}
int reset(int N,int pos)
{
    return N = N & ~(1<<pos);
}
bool check(int N,int pos)
{
    return (bool)(N & (1<<pos));
}

inline int addmod(lli x,lli y)
{
    return (x%mod + y%mod)%mod;
}
inline int submod(lli x,lli y)
{
    return (x%mod - y%mod + mod)%mod;
}
inline int mulmod(lli x,lli y)
{
    return (x%mod * y%mod)%mod;
}


inline int nextSubMask(int i, int mask)
{
    return (i-1)&mask;   /// returns next  submask
}

template<typename T>
void we_r_done(T mssg)
{
    cout<<mssg;
    exit(0);
}
const int N = 1e5 + 6;
vector<int>tree[N];
int du[N] , dv[N];

void dfsu(int x , int p){

    for(int u : tree[x]){
        if(p != u){
            du[u] = du[x] + 1;
            dfsu(u,x);
        }
    }
}

void dfsv(int x , int p){

    for(int u : tree[x]){
        if(p != u){
            dv[u] = dv[x] + 1;
            dfsv(u,x);
        }
    }
}

int main()
{
    ///freopen("output.txt","w",stdout);
   // freopen("input.txt","r",stdin);
    FastIO;

    int n,u,v;
    while(cin>>n){
        cin>>u>>v;

        for(int i=1; i<=n; i++){
            tree[i].clear();
            du[i] = dv[i] = 0;
        }
        bool bad = false;
        for(int i=1; i<n; i++){
            int x,y;
            cin>>x>>y;
            tree[x].pb(y);
            tree[y].pb(x);
        }

        du[u] = dv[v] = 0;
        dfsu(u,u);
        dfsv(v,v);
        int ans = 0;
        for(int i=1; i<=n; i++){

            ans = max(ans , dv[i]-1);

        }
        cout<<ans<<endl;
    }

    return 0;
}

///sin and cos expect input in radians not degrees. so use , sin(degrees * pi / 180)
///using bs = bitset<MX>; // how many distinct number can be form?
///sort(atoz(v), [](const data x, const data y){return (x.a==y.a?x.b>y.b : x.a<y.a);});