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
# 木によってLISを作り、dfsで抜けるときにLISをその前の状態まで復元する

import sys
sys.setrecursionlimit(1 << 25)
read = sys.stdin.readline
ra = range
enu = enumerate


def mina1(x):
    return int(x) - 1


def read_ints(mina=False):
    if mina:
        return list(map(mina1, read().split()))
    else:
        return list(map(int, read().split()))


def read_a_int():
    return int(read())


from bisect import bisect_left, bisect_right
from collections import defaultdict
N = read_a_int()
A = read_ints()
tree = defaultdict(lambda: [])
for _ in ra(N - 1):
    u, v = read_ints(mina=True)
    tree[u].append(v)
    tree[v].append(u)

LIS = []
ans = [0] * N  # 各ノードのlen(LIS)を記録


def dfs(now, p):  # 現在のノード、親
    a = A[now]
    # LISの更新
    idx = bisect_left(LIS, a)
    is_append = False
    if idx == len(LIS):
        LIS.append(a)
        is_append = True
    else:
        old = LIS[idx]   # なんの値だったか持っておく
        LIS[idx] = a  # aに更新

    ans[now] = len(LIS)  # 答えを記録
    # 次のノードを探索
    for to in tree[now]:
        if to == p:
            continue
        dfs(to, now)

    # 抜けるときにLISを復元
    if is_append:
        del LIS[idx]
    else:
        LIS[idx] = old


dfs(0, -1)
print(*ans, sep='\n')
