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.

Examples

Input

5 4 1
1 2
2 3
3 4
3 5


Output

2


Input

5 4 5
1 2
1 3
1 4
1 5


Output

1


Input

2 1 2
1 2


Output

0


Input

9 6 1
1 2
2 3
3 4
4 5
5 6
4 7
7 8
8 9


Output

5
import java.io.PrintStream;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) throws Exception{
		new Main(new Scanner(System.in), System.out, System.err).exec();
	}
	public final Scanner sc;
	public final PrintStream out, err;
	public Main(final Scanner sc, final PrintStream out, final PrintStream err) {
		this.sc = sc;this.out = out;this.err = err;
	}
	static public final long pair(int a, int b) {return (((long)a) << 32) + b;}
	static public final int getPair1(long v) {return (int)(v >> 32);}
	static public final int getPari2(long v) {return (int)(v & 0xFFFFFFFF);}

	static public final long quad(long a, long b, long c, long d) {
		return ((a&0xFFFF)<< 48) + ((b&0xFFFF)<< 32) + ((c&0xFFFF)<< 16) +(d&0xFFFF);}
	static public final int getQuad1(long v) {return (int)( v >> 48          );}
	static public final int getQuad2(long v) {return (int)((v >> 32) & 0xFFFF);}
	static public final int getQuad3(long v) {return (int)((v >> 16) & 0xFFFF);}
	static public final int getQuad4(long v) {return (int)( v        & 0xFFFF);}

	static public final long record(int e, int m, int i) {
		return quad(-1, e, m, i);
	}
	static public final int re(long record) {
		return getQuad2(record);
	}
	static public final int rm(long record) {
		return getQuad3(record);
	}
	static public final int ri(long record) {
		return getQuad4(record);
	}

	//------------------------------------------

	private int p;
	private int[] inv;
	private int[] fact;
	private int[] finv;
	public final int mod(long v) {
		v %= p;
		return (int)(v < 0 ? v+p : v);
	}
	public final int mul(long a, long b) {
		return mod(a*b);
	}
	public final void initFact(double limit, double p) {
		initFact((int)limit, (int)p);
	}
	public final void initFact(int limit, int p) {
		this.p = p;
		this.inv = new int[limit+1];
		this.fact = new int[limit+1];
		this.finv = new int[limit+1];
		fact[1] = finv[1] = inv[0] = inv[1] = 1;
		for(int i=2;i<=limit;i++) {
			int a = p / i;
			int b = p % i;
			inv[i]  = mul(-a, inv[b]);
			fact[i] = mul(fact[i-1], i);
			finv[i] = mul(finv[i-1], inv[i]);
		}
	}
	public final int comb(int m, int n) {
		return mul(mul(fact[m], finv[n]), finv[m-n]);
	}
	//------------------------------------------
	static long lcm (long a, long b) {
		long temp;
		long c = a;
		c *= b;
		while((temp = a%b)!=0) {
			a = b;
			b = temp;
		}
		return (long)(c/b);
	}

	//------------------------------------------


	static public class Node{
		private final LinkedList<Node> next = new LinkedList<>();
		private int u = -1;
		private int v = -1;
		public boolean isEnd() {
			// uがここに到着するまでの回数がvと同じか、より後の場合、
			// ここに来るかここでつかまることになる。
			return u >= v;
		}
		public int diff() {
			return v - u;
		}
	}

	public void exec() throws Exception{
		final int N = sc.nextInt();
		final int u = sc.nextInt();
		final int v = sc.nextInt();
		final Node[] graph = new Node[N+1];
		for(int i=1;i<=N;i++) {
			graph[i] = new Node();
		}
		for(int i=1;i<=N-1;i++) {
			final int A = sc.nextInt();
			final int B = sc.nextInt();
			graph[A].next.add(graph[B]);
			graph[B].next.add(graph[A]);
		}
		final ArrayList<Node> list = new ArrayList<>(N);
		list.add(graph[v]);
		graph[v].v=0;
		while(!list.isEmpty()) {
			final Node node = list.remove(list.size()-1);
			for(Node next : node.next) {
				if(next.v==-1) {
					next.v = node.v+1;
					list.add(next);
				}
			}
		}
		list.add(graph[u]);
		graph[u].u = 0;
		Node max = graph[u];
		while(!list.isEmpty()) {
			final Node node = list.remove(list.size()-1);
			if(node.isEnd()){
				if(max.diff() < node.diff()) {
					max = node;
				}
				continue;
			}
			boolean isEnd = true;
			for(Node next : node.next) {
				if(next.u == -1) {
					next.u = node.u+1;
					list.add(next);
					isEnd = false;
				}
			}
			// 終端に来ている場合
			if(isEnd) {
				if(max.v < node.v) {
					max = node;
				}
			}
		}
		out.println(max.v-1);
	}
}
