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)

#define _CRT_SECURE_NO_WARNINGS

#include <iostream>
#include <complex>
#include <chrono>
#include <random>
#include <numeric>
#include <queue>
#include <map>
#include <list>
#include <vector>
#include <string>
#include <stack>
#include <limits>
#include <climits>
#include <cassert>
#include <fstream>
#include <cstring>
#include <cmath>
#include <bitset>
#include <iomanip>
#include <algorithm>
#include <functional>
#include <cstdio>
#include <ciso646>
#include <set>
#include <array>
#include <unordered_map>
#include <unordered_set>
#include <type_traits>

using namespace std;

#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define RFOR(i, a, b) for (int i = (b)-1; i >= (a); i--)
#define REP(i, n) for (int i = 0; i < (n); i++)
#define RREP(i, n) for (int i = (n)-1; i >= 0; i--)
#define inf 0x3f3f3f3f3f3f3f3f
#define ALL(a) (a).begin(), (a).end()
#define DEBUG(x) // cerr<<#x<<": "<<x<<endl
#define ll long long
#define ull unsigned long long
using pii = pair<ll, ll>;
#define eps 1e-14
#define SETUP cin.tie(0), ios::sync_with_stdio(false), cout << setprecision(15) << std::fixed;
template<class T>bool chmax(T& a, const T& b) { if (a < b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T& a, const T& b) { if (b < a) { a = b; return 1; } return 0; }

template <class T>
using vec2 = std::vector<vector<T>>;

//// bit ////
#ifdef _MSC_VER
#ifdef _WIN32
inline unsigned int __builtin_ctz(unsigned int x) { unsigned long r; _BitScanForward(&r, x); return r; }
inline unsigned int __builtin_clz(unsigned int x) { unsigned long r; _BitScanReverse(&r, x); return 31 - r; }
inline unsigned int __builtin_ffs(unsigned int x) { unsigned long r; return _BitScanForward(&r, x) ? r + 1 : 0; }
// inline unsigned int __builtin_popcount(unsigned int x){ return __popcnt(x); }
#ifdef _WIN64
inline unsigned long long __builtin_ctzll(unsigned long long x) { unsigned long r; _BitScanForward64(&r, x); return r; }
inline unsigned long long __builtin_clzll(unsigned long long x) { unsigned long r; _BitScanReverse64(&r, x); return 63 - r; }
inline unsigned long long __builtin_ffsll(unsigned long long x) { unsigned long r; return _BitScanForward64(&r, x) ? r + 1 : 0; }
inline unsigned long long __builtin_popcountll(unsigned long long x) { return __popcnt64(x); }
#else
inline unsigned int hidword(unsigned long long x) { return static_cast<unsigned int>(x >> 32); }
inline unsigned int lodword(unsigned long long x) { return static_cast<unsigned int>(x & 0xFFFFFFFF); }
inline unsigned long long __builtin_ctzll(unsigned long long x) { return lodword(x) ? __builtin_ctz(lodword(x)) : __builtin_ctz(hidword(x)) + 32; }
inline unsigned long long __builtin_clzll(unsigned long long x) { return hidword(x) ? __builtin_clz(hidword(x)) : __builtin_clz(lodword(x)) + 32; }
inline unsigned long long __builtin_ffsll(unsigned long long x) { return lodword(x) ? __builtin_ffs(lodword(x)) : hidword(x) ? __builtin_ffs(hidword(x)) + 32 : 0; }
// inline unsigned long long __builtin_popcountll(unsigned long long x) { return __builtin_popcount(lodword(x)) + __builtin_popcount(hidword(x)); }
#endif // _WIN64
#endif // _WIN32
#endif // _MSC_VER

namespace
{
	struct input_returnner
	{
		ll N;
		input_returnner(ll N_ = 0) : N(N_) {}
		template <typename T>
		operator vector<T>() const
		{
			vector<T> res(N);
			for (auto& a : res)
				cin >> a;
			return std::move(res);
		}
		template <typename T>
		operator T() const
		{
			T res;
			cin >> res;
			return res;
		}
		template <typename T>
		T operator-(T right) { return T(input_returnner()) - right; }
		template <typename T>
		T operator+(T right) { return T(input_returnner()) + right; }
		template <typename T>
		T operator*(T right) { return T(input_returnner()) * right; }
		template <typename T>
		T operator/(T right) { return T(input_returnner()) / right; }
		template <typename T>
		T operator<<(T right) { return T(input_returnner()) << right; }
		template <typename T>
		T operator>>(T right) { return T(input_returnner()) >> right; }
	};
	template <typename T>
	input_returnner in() { return in<T>(); }
	input_returnner in() { return input_returnner(); }
	input_returnner in(ll N) { return std::move(input_returnner(N)); }
} // namespace

template <typename T>
istream& operator>>(istream& is, vector<T>& vec)
{
	for (T& x : vec)
		is >> x;
	return is;
}

template <typename T>
struct is_vector : std::false_type
{
};

template <typename T>
struct is_vector<std::vector<T>> : std::true_type
{
};

template <typename T>
constexpr bool is_vector_v = is_vector<T>::value;

template <typename T>
std::ostream& operator<<(std::ostream& out, const std::vector<T>& v)
{
	if (!v.empty())
	{
		for (int i = 0; i < v.size(); ++i)
		{
			out << v[i] << (i == v.size() - 1 ? "\n" : (is_vector_v<T> ? "" : ", "));
		}
	}
	return out;
}

namespace std
{
	// ref: https://stackoverflow.com/questions/7110301/generic-hash-for-tuples-in-unordered-map-unordered-set
	template <class T>
	inline void hash_combine(std::size_t& seed, T const& v)
	{
		seed ^= std::hash<T>()(v) + 0x9e3779b9 + (seed << 6) + (seed >> 2);
	}

	// Recursive template code derived from Matthieu M.
	template <class Tuple, size_t Index = std::tuple_size<Tuple>::value - 1>
	struct HashValueImpl
	{
		static void apply(size_t& seed, Tuple const& tuple)
		{
			HashValueImpl<Tuple, Index - 1>::apply(seed, tuple);
			hash_combine(seed, std::get<Index>(tuple));
		}
	};

	template <class Tuple>
	struct HashValueImpl<Tuple, 0>
	{
		static void apply(size_t& seed, Tuple const& tuple)
		{
			hash_combine(seed, std::get<0>(tuple));
		}
	};
	template <typename... TT>
	struct hash<std::tuple<TT...>>
	{
		size_t operator()(std::tuple<TT...> const& tt) const
		{
			size_t seed = 0;
			HashValueImpl<std::tuple<TT...>>::apply(seed, tt);
			return seed;
		}
	};

	template <class T, class U>
	class hash<std::pair<T, U>>
	{
	public:
		size_t operator()(const std::pair<T, U>& x) const
		{
			return hash<std::tuple<T, U>>()(std::tie(x.first, x.second));
		}
	};
} // namespace std

// ref: https://stackoverflow.com/questions/6245735/pretty-print-stdtuple
namespace aux
{
	template <std::size_t...>
	struct seq
	{
	};

	template <std::size_t N, std::size_t... Is>
	struct gen_seq : gen_seq<N - 1, N - 1, Is...>
	{
	};

	template <std::size_t... Is>
	struct gen_seq<0, Is...> : seq<Is...>
	{
	};

	template <class Ch, class Tr, class Tuple, std::size_t... Is>
	void print_tuple(std::basic_ostream<Ch, Tr>& os, Tuple const& t, seq<Is...>)
	{
		using swallow = int[];
		(void)swallow {
			0, (void(os << (Is == 0 ? "" : ", ") << std::get<Is>(t)), 0)...
		};
	}
} // namespace aux

template <class Ch, class Tr, class... Args>
auto operator<<(std::basic_ostream<Ch, Tr>& os, std::tuple<Args...> const& t)
-> std::basic_ostream<Ch, Tr> &
{
	os << "(";
	aux::print_tuple(os, t, aux::gen_seq<sizeof...(Args)>());
	return os << ")";
}

template <class S, class T>
std::ostream& operator<<(std::ostream& os, const std::pair<S, T>& p)
{
	return os << "(" << p.first << ", " << p.second << ")";
}

// ref: https://stackoverflow.com/questions/8542591/c11-reverse-range-based-for-loo�Fp
template <typename T>
struct reversion_wrapper
{
	T& iterable;
};

template <typename T>
auto begin(reversion_wrapper<T> w) { return std::rbegin(w.iterable); }

template <typename T>
auto end(reversion_wrapper<T> w) { return std::rend(w.iterable); }

template <typename T>
reversion_wrapper<T> REV(T&& iterable) { return { iterable }; }

template <class T>
bool inside(T left, T val, T right)
{
	return left <= val and val < right;
}

template <class T>
T bitCount(T num)
{
	T res = 0;
	while (num > 0)
	{
		if (num & 1)
			++res;
		num >>= 1;
	}
	return res;
}

ll MOD = 1e9 + 7;
//ll MOD = 998244353;

void solve();

signed main()
{
	SETUP;
#ifdef _DEBUG
	while (true)
	{
#endif
		solve();
#ifdef _DEBUG
		cout << "-------" << endl;
	}
#endif
#ifdef _DEBUG
	system("pause");
#endif
	return 0;
}

#define int ll

struct mint {
	int64_t value;
	mint() :value(0) {}
	mint(int64_t value_) : value(value_% MOD) {}
	inline mint& operator += (const mint& rhs) {
		value += rhs.value;
		if (value >= MOD) { value -= MOD; }
		return *this;
	}
	inline mint& operator -= (const mint& rhs) {
		value -= rhs.value;
		if (value < 0) {
			value += MOD;
		}
		return *this;
	}
	inline mint& operator *= (const mint& rhs) {
		value *= rhs.value;
		value %= MOD;
		return *this;
	}
	inline mint& operator %= (const mint& rhs) {
		return *this;
	}
};

// x^n % mod
// O(log n)
mint mod_pow(mint x, uint64_t n) {
	mint res = 1;
	while (n > 0) {
		if (n & 1) res *= x;
		x *= x;
		n >>= 1;
	}
	return res;
}

string bin(int x) {
	string res = "";
	RREP(i, 32) {
		res += ((x >> i) & 1) ? "1" : "0";
	}
	return res;
}

inline mint operator + (const mint& lhs, const mint& rhs) { mint res = lhs; res += rhs; return res; }
inline mint operator - (const mint& lhs, const mint& rhs) { mint res = lhs;	res -= rhs;	return res; }
inline mint operator * (const mint& lhs, const mint& rhs) { mint res = lhs;	res *= rhs;	return res; }
inline mint operator % (const mint& lhs, const uint64_t& rhs) { return lhs; }

ostream& operator << (ostream& out, mint n) { return out << n.value; }

int gcd(int a, int b) {
	return b != 0 ? gcd(b, a % b) : a;
}

void DFS(vector<vector<int>>& G, int node, int depth, vector<int>& res) {
	res[node] = depth;
	for (auto& dst : G[node]) {
		if(res[dst] == -1) DFS(G, dst, depth+1, res);
	}
}

int solve(vector<vector<int>>& G, int node, int depth, vector<int>& Tak, vector<int>& Ao, vector<bool>& memo) {
	memo[node] = true;
	int res = -1;
	for (auto& dst : G[node]) {
		if(not memo[dst] and Tak[node] < Ao[node])
			res = max(solve(G, dst, depth+1, Tak, Ao, memo), res);
	}
	int a = Ao[node];
	--a;
	//cerr << "node:" << node << " out:" << a << endl;
	res = max(res, a);
	return res;
}

void solve() {
	int N, u, v; cin >> N >> u >> v;
	--u, --v;

	vector<vector<int> > G(N);
	REP(i, N - 1) {
		int a, b; cin >> a >> b;
		--a, --b;
		G[a].emplace_back(b);
		G[b].emplace_back(a);
	}

	vector<int> TakLength(N, -1);
	DFS(G, u, 0, TakLength);

	vector<int> AoLength(N, -1);
	DFS(G, v, 0, AoLength);

	vector<bool> memo(N);
	int res = solve(G, u, 0, TakLength, AoLength, memo);
	cout << res << endl;
}