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 Kotlin (1.0.0) (judge verdict: Accepted)

import java.io.*
import java.util.*

val pw = PrintWriter(System.out)
const val MOD = 1000000007L
const val INF = 2147483647
const val LINF = 9223372036854775807L

fun main(args: Array<String>) {
    solve()
    pw.flush()
}

fun solve() {
    val (n, takahashi, aoki) = nextIntList()
    val nodes = Array(n) { Node(it) }
    repeat(n - 1){
        val (a, b) = nextIntList()
        nodes[a - 1].edges.add(Edge(a - 1, b - 1, 1))
        nodes[b - 1].edges.add(Edge(b - 1, a - 1, 1))
    }

    val tdist = bfs(nodes, takahashi - 1)
    val adist = bfs(nodes, aoki - 1)

    var ans = 0L
    for (i in 0 until n){
        val t = tdist[i]
        val a = adist[i]
        if(t >= a) continue
        val res = a - 1
        ans = max(ans, res)
    }

    println(ans)

}

fun bfs(nodes: Array<Node>, start: Int) : Array<Long> {
    val queue = ArrayDeque<Int>()
    queue.add(start)
    val dist: Array<Long> = Array(nodes.size) { -1L }
    dist[start] = 0
    while(queue.isNotEmpty()){
        val now = queue.poll()
        for (edge in nodes[now].edges) {
            if(dist[edge.to] != -1L) continue
            dist[edge.to] = dist[now] + 1
            queue.add(edge.to)
        }
    }
    return dist
}

data class Node(val id: Int, var past: Int = -1, var edges: MutableList<Edge> = mutableListOf())
data class Edge(val from: Int, val to: Int, val cost: Long = 1L)

fun println(v: String) {
    pw.println(v)
}

fun print(v: String) {
    pw.print(v)
}

fun next() = readLine()!!
fun nextInt() = next().toInt()
fun nextLong() = next().toLong()
fun nextDouble() = next().toDouble()
fun nextList() = next().split(" ").toMutableList()
fun nextIntList(index: Int = 0) = next().split(" ").map { it.toInt() + index }.toMutableList()
fun nextLongList(index: Long = 0L) = next().split(" ").map { it.toLong() + index }.toMutableList()
fun nextDoubleList() = next().split(" ").map { it.toDouble() }.toMutableList()

fun ary(n: Int, init: String = "") = Array(n) { init }
fun intAry(n: Int, init: Int = 0) = IntArray(n) { init }
fun longAry(n: Int, init: Long = 0L) = LongArray(n) { init }
fun doubleAry(n: Int, init: Double = 0.0) = DoubleArray(n) { init }
fun ary2(n: Int, m: Int, init: String = "") = Array(n) { Array(m) { init } }
fun intAry2(n: Int, m: Int, init: Int = 0) = Array(n) { IntArray(m) { init } }
fun longAry2(n: Int, m: Int, init: Long = 0) = Array(n) { LongArray(m) { init } }
fun doubleAry2(n: Int, m: Int, init: Double = 0.0) = Array(n) { DoubleArray(m) { init } }
fun ary3(n: Int, m: Int, k: Int, init: String = "") = Array(n) { Array(m) { Array(k) { init } } }
fun intAry3(n: Int, m: Int, k: Int, init: Int = 0) = Array(n) { Array(m) { IntArray(k) { init } } }
fun longAry3(n: Int, m: Int, k: Int, init: Long = 0L) = Array(n) { Array(m) { LongArray(k) { init } } }
fun doubleAry3(n: Int, m: Int, k: Int, init: Double = 0.0) = Array(n) { Array(m) { DoubleArray(k) { init } } }

fun abs(n: Int): Int = Math.abs(n)
fun abs(n: Long): Long = Math.abs(n)
fun abs(n: Double): Double = Math.abs(n)
fun max(vararg values: Int) = if (values.isEmpty()) -INF else values.max()!!
fun max(vararg values: Long) = if (values.isEmpty()) -LINF else values.max()!!
fun min(vararg values: Int) = if (values.isEmpty()) INF else values.min()!!
fun min(vararg values: Long) = if (values.isEmpty()) LINF else values.min()!!