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 Python (3.8.2) (judge verdict: Runtime Error)

# LIS on Tree

from collections import deque
import bisect

N = int(input())
V = list(map(int, input().split()))
V.insert(0, 0)
visited = [False for i in range(N+1)]
inf = 10**18
dp = [inf for i in range(N+1)]
ans = [-1 for i in range(N+1)]
stack = deque()

G = [[] for i in range(N+1)]
for _ in range(N-1):
    u, v = map(int, input().split())
    G[u].append(v)
    G[v].append(u)


def dfs(s_node):
    visited[s_node] = True
    # dp値の更新
    value = V[s_node]
    index = bisect.bisect_left(dp, value)
    old_value = dp[index]
    dp[index] = value
    ans[s_node] = bisect.bisect_right(dp, inf-1)
    stack.append((index, old_value, value))
    for child in G[s_node]:
        if visited[child] == False:
            dfs(child)
    # 帰りがけにバックトラックを行う
    index, old_value, value = stack.pop()
    dp[index] = old_value


dfs(1)

for i in range(1, N+1):
    print(ans[i])