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)

#!/usr/bin/python3

import sys
from bisect import bisect_left

sys.setrecursionlimit(1000000)

n = int(input())
a = list(map(int, input().split()))

nbs = [ [] for _ in range(n) ]

for _ in range(n - 1):
    (u, v) = map(int, input().split())
    nbs[u - 1].append(v - 1)
    nbs[v - 1].append(u - 1)


mlen =  [ 0 ] * n
dp = [ float('inf') ] * n

def getmlen(cur, p, ctail):
    for ch in nbs[cur]:
        if ch == p:
            continue

        idx = bisect_left(dp, a[ch])
        ov = dp[idx]

        ntail = max(ctail, idx)
        dp[idx] = a[ch]
        mlen[ch] = ntail + 1
        getmlen(ch, cur, ntail)
        dp[idx] = ov

        

dp[0] = a[0]
mlen[0] = 1
getmlen(0, -1, 0)

for i in range(n):
    print(mlen[i])