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 Python (3.4.3) (judge verdict: Runtime Error)

# 解説を見た

def solve(N, t, a, g) -> int:
    def calc_dist(init):
        dist = [-1] * N
        dist[init] = 0

        def _calc_dist(curr):
            for next_ in g[curr]:
                if dist[next_] != -1:
                    continue
                dist[next_] = dist[curr] + 1
                _calc_dist(next_)

        _calc_dist(init)
        return dist

    dist_a = calc_dist(a)
    dist_t = calc_dist(t)

    return max(dist_a[v] for v in range(N) if dist_t[v] < dist_a[v]) - 1


if __name__ == '__main__':
    N, t, a = map(int, input().split())
    t -= 1
    a -= 1

    g = tuple(set() for _ in range(N))
    for _ in range(N - 1):
        u, v = (int(x) - 1 for x in input().split())
        g[u].add(v)
        g[v].add(u)

    print(solve(N, t, a, g))

# 青木君からの距離 > 高橋君からの距離 となる頂点Xが目的地の候補
# 青木君と高橋君の共通祖先からの距離は等しいので、
# 共通祖先への距離は、 青木君 > 高橋君 となり、高橋君は必ずXに到達できる
# Xに着いた高橋君は、手前の頂点とXとを行ったり来たりする
# a-t->at-: aが一歩進んだときに高橋君に会う(-は無人のマス目)
# at: tが手前の頂点に移動したときに青木君に会う
# いずれも手前の頂点まで青木君が移動したときにゲームが終わる
# どのXを選んでも、Xの手前でゲームが終わるので、青木君から最も遠いXを選ぶのが最善
# 頂点X: 追いつめられる位置=葉