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.

Example

Input

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


Output

1
2
3
3
4
4
5
2
2
3
from bisect import bisect_right
from sys import setrecursionlimit

setrecursionlimit(10**8)
n = int(input())
A = list(map(int, input().split()))
gr = []
for i in range(n):
    gr.append([])
for i in range(n - 1):
    a, b = map(int, input().split())
    a -= 1
    b -= 1
    gr[a].append(b)
    gr[b].append(a)
ans = [0] * n
dp = [10 ** 10 for i in range(1 + n)]
dp[0] = -10 ** 10
kek = 1
used = {0}


def dfs(v):
    global kek
    j = bisect_right(dp, A[v] - 1)
    old = dp[j]
    if dp[j - 1] < A[v] < dp[j]:
        old = dp[j]
        dp[j] = A[v]
        if kek == j:
            for h in range(kek + 1, 1 + n):
                if dp[h] == 10 ** 10:
                    kek = h
                    break
            if kek == j:
                kek = n + 1
    ans[v] = kek - 1
    for u in gr[v]:
        if u not in used:
            used.add(u)
            dfs(u)
    dp[j] = old
    if kek > j and old == 10 ** 10:
        kek = j


dfs(0)
print(*ans)