// Original C++ solution (slower):
#include <bits/stdc++.h>



using namespace std;



typedef long long int ll;

typedef pair<ll, ll> pll;



#define FOR(i, n, m) for(ll (i)=(m);(i)<(n);++(i))

#define REP(i, n) FOR(i,n,0)

#define OF64 std::setprecision(10)



const ll MOD = 1000000007;

const ll INF = (ll) 1e15;



struct Vertex {

    vector<ll> node;

};



Vertex V[100005];

ll T[100005];

ll A[100005];



int main() {

    ll N, u, v;

    cin >> N >> u >> v;

    u--;

    v--;

    REP(i, N - 1) {

        ll a, b;

        cin >> a >> b;

        a--;

        b--;

        V[a].node.push_back(b);

        V[b].node.push_back(a);

    }



    {

        memset(A, -1, sizeof(A));

        A[v] = 0;

        queue<pll> q;

        q.push(pll(v, 0));

        while (!q.empty()) {

            pll t = q.front();

            q.pop();

            REP(i, V[t.first].node.size()) {

                ll n = V[t.first].node[i];

                if (A[n] >= 0)

                    continue;

                A[n] = t.second + 1;

                q.push(pll(n, t.second + 1));

            }

        }

    }



    ll max = 0;

    {

        memset(T, -1, sizeof(T));

        T[u] = 0;

        queue<pll> q;

        q.push(pll(u, 0));

        while (!q.empty()) {

            pll t = q.front();

            q.pop();

            if (V[t.first].node.size() > 1) {

                max = std::max(max, A[t.first]);

            }

            REP(i, V[t.first].node.size()) {

                ll n = V[t.first].node[i];

                if (T[n] >= 0)

                    continue;

                T[n] = t.second + 1;

                if (T[n] <= A[n])

                    q.push(pll(n, t.second + 1));

            }

        }

    }



    cout << max << endl;

    return 0;

}

// Optimized solution to the same problem (faster):
#include <bits/stdc++.h>



using namespace std;



typedef long long int ll;

typedef pair<ll, ll> pll;



#define FOR(i, n, m) for(ll (i)=(m);(i)<(n);++(i))

#define REP(i, n) FOR(i,n,0)

#define OF64 std::setprecision(10)



const ll MOD = 1000000007;

const ll INF = (ll) 1e15;



struct RootedTree {

    RootedTree(int n, int root = 0) :

            mRoot(root) {

        mVertexTbl.resize(n);

    }



    void addEdge(int a, int b, long long int cost = 1) {

        mVertexTbl[a].child.push_back(pair<int, long long int>(b, cost));

        mVertexTbl[b].child.push_back(pair<int, long long int>(a, cost));

    }



    void build() {

        queue<tuple<int, int, long long int>> q;

        q.push(make_tuple(mRoot, -1, 0));

        while (!q.empty()) {

            auto t = q.front();

            q.pop();

            int index = get<0>(t);

            int parent = get<1>(t);

            long long int depth = get<2>(t);

            mVertexTbl[index].parent = parent;

            mVertexTbl[index].depth = depth;

            for (auto it = mVertexTbl[index].child.begin(); it != mVertexTbl[index].child.end();) {

                int nxt = (*it).first;

                if (nxt == parent) {

                    mVertexTbl[index].child.erase(it);

                    continue;

                }

                q.push(make_tuple(nxt, index, depth + (*it).second));

                ++it;

            }

        }

    }



    struct Vertex {

        vector<pair<int, long long int>> child;

        long long int depth = 0;

        int parent = -1;

    };



    int mRoot;

    vector<Vertex> mVertexTbl;

};



ll CA[100005];

ll CT[100005];



int main() {

    cin.tie(0);

    ios::sync_with_stdio(false);

    ll N, T, A;

    cin >> N >> T >> A;

    T--;

    A--;

    RootedTree tree(N, A);

    REP(i, N - 1) {

        ll a, b;

        cin >> a >> b;

        a--;

        b--;

        tree.addEdge(a, b);

    }

    tree.build();

    REP(i, N) {

        CA[i] = CT[i] = INF;

    }



    {

        queue<pll> q;

        q.push(pll(A, 0));

        CA[A] = 0;

        while (!q.empty()) {

            pll t = q.front();

            q.pop();

            REP(i, tree.mVertexTbl[t.first].child.size()) {

                ll n = tree.mVertexTbl[t.first].child[i].first;

                if (CA[n] <= t.second + 1)

                    continue;

                CA[n] = t.second + 1;

                q.push(pll(n, CA[n]));

            }

        }

    }

    {

        queue<pll> q;

        q.push(pll(T, 0));

        CT[T] = 0;

        while (!q.empty()) {

            pll t = q.front();

            q.pop();

            ll cost = t.second + 1;

            ll n = tree.mVertexTbl[t.first].parent;

            if (CT[n] <= cost)

                continue;

            if (CA[n] <= cost)

                continue;

            CT[n] = cost;

            q.push(pll(n, cost));

        }

    }



    ll ta = T;

    REP(i, N) {

        if (CT[i] == INF)

            continue;

        if (CT[ta] < CT[i])

            ta = i;

    }

    ll ans = CT[ta];

    {

        queue<pll> q;

        q.push(pll(ta, CT[ta]));

        while (!q.empty()) {

            pll t = q.front();

            q.pop();

            ll cost = t.second + 1;

            ans = std::max(t.second, ans);

            REP(i, tree.mVertexTbl[t.first].child.size()) {

                ll n = tree.mVertexTbl[t.first].child[i].first;

                if (CA[n] >= cost)

                    ans = std::max(ans, std::min(CA[n], cost));

                if (CA[n] > cost)

                    q.push(pll(n, cost));

            }

            ll p = tree.mVertexTbl[t.first].parent;

            if (CA[p] >= cost)

                ans = std::max(ans, std::min(CA[p], cost));

            if (CA[p] > cost)

                q.push(pll(p, cost));

        }

    }

    cout << ans << endl;

    return 0;

}

// Fastest known solution for this problem: