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_left
import sys
sys.setrecursionlimit(10**7)
# sys.stdin.readline

# スペース区切りの入力を読み込んで数値リストにして返します。
def get_nums_l():
    return [ int(s) for s in input().split(" ")]

def log(*args):
    print("DEBUG:", *args, file=sys.stderr)

INF = 999999999999999999999999

n = int(input())
A = [-INF] + get_nums_l()

edges = [ [] for _ in range(n+1) ]
for line in map(lambda s: s.strip(), sys.stdin.readlines()):
    u,v = map(int, line.split())
    edges[u].append(v)
    edges[v].append(u)

def dfs(dp, u, p=None):

    # log(u)
    # log(dp)

    # dpの中でA[u]以上の値が入っている最小のindex
    i = bisect_left(dp, A[u])

    old = dp[i]
    dp[i] = A[u]

    # 有効値(INF以外)が入っている最大のindex
    ans[u] = bisect_left(dp, INF) - 1

    for v in edges[u]:
        if v == p:
            continue
        dfs(dp, v, u)

    # 巻き戻し
    dp[i] = old

# ans[i] = ノード1からノードiへの経路中の最長増加部分列長
ans = [0] * (n+1)

# dp[i] = 最長増加部分列長がiの場合の末尾ノード値の最小
# dp[0]は使用しないので負の無限大を入れておく
dp = [INF] * (n+1)
dp[0] = -INF

dfs(dp, 1, None)

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