Problem:
Score : 600 points

Problem Statement

We have a tree with N vertices, whose i-th edge connects Vertex u_i and Vertex v_i.
Vertex i has an integer a_i written on it.
For every integer k from 1 through N, solve the following problem:

We will make a sequence by lining up the integers written on the vertices along the shortest path from Vertex 1 to Vertex k, in the order they appear. Find the length of the longest increasing subsequence of this sequence.

Here, the longest increasing subsequence of a sequence A of length L is the subsequence A_{i_1} , A_{i_2} , ... , A_{i_M} with the greatest possible value of M such that 1 \leq i_1 < i_2 < ... < i_M \leq L and A_{i_1} < A_{i_2} < ... < A_{i_M}.

Constraints

2 \leq N \leq 2 \times 10^5

1 \leq a_i \leq 10^9

1 \leq u_i , v_i \leq N

u_i \neq v_i

The given graph is a tree.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N
u_1 v_1
u_2 v_2
:
u_{N-1} v_{N-1}

Output

Print N lines. The k-th line, print the length of the longest increasing subsequence of the sequence obtained from the shortest path from Vertex 1 to Vertex k.

Sample Input 1

10
1 2 5 3 4 6 7 3 2 4
1 2
2 3
3 4
4 5
3 6
6 7
1 8
8 9
9 10

Sample Output 1

1
2
3
3
4
4
5
2
2
3

For example, the sequence A obtained from the shortest path from Vertex 1 to Vertex 5 is 1,2,5,3,4. Its longest increasing subsequence is A_1, A_2, A_4, A_5, with the length of 4.

Solution in PyPy3 (7.3.0) (judge verdict: Time Limit Exceeded)

import sys
from bisect import bisect_left, bisect_right
sys.setrecursionlimit(10**6)
def input(): return sys.stdin.readline().strip()


def main():
    N = int(input())
    A = list(map(int, input().split()))
    repn = [[] for _ in range(N)]
    for _ in range(N - 1):
        u, v = map(int, input().split())
        repn[u - 1].append(v - 1)
        repn[v - 1].append(u - 1)
    
    length = [0] * (N + 1)
    dp = [-1]

    """
    やはり各頂点でLIS配列を取ると配列コピーの時間がかかってTLEするので、
    配列は唯一つを扱って、DFSからバックするときに配列も元に戻す操作を行いたい
    （ここの実装の仕方がわからなかった）

    これはdfsを再帰で書いた後に復元操作を追記すれば異常に楽に書ける。。。
    """

    def dfs(v, p=N):
        if A[v] > dp[-1]:
            dp.append(A[v])
            length[v] = length[p] + 1
            for u in repn[v]:
                if u == p: continue
                dfs(u, v)
            dp.pop()
        else:
            length[v] = length[p]
            idx = bisect_left(dp[:length[p]], A[v]) # bisect_rightだとdpに重複する値を書き込んでしまうのでダメ
            old = dp[idx]
            dp[idx] = A[v]
            for u in repn[v]:
                if u == p: continue
                dfs(u, v)
            dp[idx] = old
        #print("v={}, len={}, dp={}".format(v, length[v], dp))

    dfs(0)
    length.pop()
    for ans in length:
        print(ans)
    


if __name__ == "__main__":
    main()