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 int long long int
#define pll pair<ll,ll>
#define vi vector < int >
#define pii pair<int,int>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define all(a) (a).begin(),(a).end()
#define s(x) scanf("%d",&x)
#define sll(x) scanf("%lld",&x)
#define pf(x) printf("%d\n",x)
#define pfl(x) printf("%lld\n",x)
#define flash ios_base::sync_with_stdio(false);cin.tie(0)
#define goog(x) cout << "Case #" << x << ": "
#define MAX5 200005
#define MAX6 2000005
#define MLOG 18
#define limit 1e18
#define mod 1000000007   
void __print(int x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}
 
template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifndef ONLINE_JUDGE
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif
 
 
int power(int x, int y, int p){
    int res = 1;
    x = x % p;
    while (y > 0){
        if (y & 1)
            res = (res*x) % p;
        y = y>>1;
        x = (x*x) % p;
    }
    return res;
}
vector < int > v[MAX5];
int ans1[MAX5], ans2[MAX5];
 
int bfs1(int u){
    queue < int > q;
    q.push(u);
    ans1[u] = 0;
    while(!q.empty()){
        int x = q.front();
        q.pop();
        for(auto e : v[x]){
            if(ans1[e] > ans1[x] + 1){
                ans1[e] = ans1[x] + 1;
                q.push(e);
            }
        }
    }
}   
int bfs2(int u){
    queue < int > q;
    q.push(u);
    ans2[u] = 0;
    while(!q.empty()){
        int x = q.front();
        q.pop();
        for(auto e : v[x]){
            if(ans2[e] > ans2[x] + 1){
                ans2[e] = ans2[x] + 1;
                q.push(e);
            }
        }
    }
}   
void solve(){
    int n, u, ff;
    cin >> n >> u >> ff;
    if(n == 2){
        cout << 0 << endl;
        return;
    }
    for(int i = 1; i <= n; i++) ans1[i] = limit;
    for(int i = 1; i <= n; i++) ans2[i] = limit;
    for(int i = 0; i < n-1; i++){
        int a, b;
        cin >> a >> b;
        v[a].pb(b);
        v[b].pb(a);
    }
    bfs1(u);
    bfs2(ff);
    int ans = 0, aa = limit;
    for(int i = 1; i <= n; i++){
        if(ans2[i] > ans1[i]){
            ans = max(ans, ans2[i]);
        }
        //debug(ans1[i] , ans2[i]);
    }
    if(aa == limit) aa = -1;
    cout << ans-1;
} 
signed main(){  
//  freopen("ii.txt", "w", stdin);
    flash;
    int t = 1;
    //  cin >> t;  
    for(int tc = 1; tc <= t; tc++){
    //goog(tc);
        solve();
    }
}