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)

from bisect import bisect_left
import sys
sys.setrecursionlimit(1000000)


def dfs(v, p):
    i = bisect_left(dp, a[v])
    dp[i], old = a[v], dp[i]
    ans[v] = bisect_left(dp, INF)
    for u in to[v]:
        if u == p:
            continue
        dfs(u, v)
    dp[i] = old

if __name__ == "__main__":
    n = int(input())
    a = list(map(int, input().split()))
    to = [[] for _ in range(n)]
    ans = [0]*n
    INF = 10**10
    for _ in range(n-1):
        u, v = map(int, input().split())
        u -= 1
        v -= 1
        to[u].append(v)
        to[v].append(u)
    dp = [INF]*n
    dp[0] = -INF
    dfs(0, -1)
    print(*ans, sep="\n")