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
import math
from decimal import Decimal, ROUND_HALF_UP, ROUND_HALF_EVEN
from collections import deque
def I(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LI2(N): return [list(map(int, sys.stdin.readline().split())) for i in range(N)]
def S(): return sys.stdin.readline().rstrip()
def LS(): return sys.stdin.readline().split()
def LS2(N): return [sys.stdin.readline().split() for i in range(N)]
def FILL(i,h): return [i for j in range(h)]
def FILL2(i,h,w): return [FILL(i,w) for j in range(h)]
def FILL3(i,h,w,d): return [FILL2(i,w,d) for j in range(h)]
def FILL4(i,h,w,d,d2): return [FILL3(i,w,d,d2) for j in range(h)]
def sisha(num,digit): return Decimal(str(num)).quantize(Decimal(digit),rounding=ROUND_HALF_UP)
#'0.01'や'1E1'などで指定、整数に戻すならintをかます
MOD = 1000000007
INF = float("inf")
sys.setrecursionlimit(10**5+10)
#input = sys.stdin.readline
from bisect import bisect_left

def dfs(i,before):
    global seq
    global ans
    added  = 0

    #現在地のAの値を、以前までのseqリストのどこに追加するか決める
    pos = bisect_left(seq,a[i-1])
    if len(seq)<= pos:
        seq.append(a[i-1])
        added  = 1
    else:
        old = seq[pos]
        seq[pos]=a[i-1]
    ans[i-1]=len(seq)

    #隣接する頂点に関して再帰
    for u in to[i]:
        if u==before:
            continue
        dfs(u,i)

    #seq配列をもとに戻す
    if added==1:
        seq=seq[:-1]
    else:
        seq[pos]=old


N = I()
a = LI()
to = [[] for i in range(N+1)]
to[0] += [1]
for i in range(N-1):
    u,v = MI()
    to[u].append(v)
    to[v].append(u)
seq = []
ans = [-1]*(N)

dfs(1,-1)
[print(i) for i in ans]