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
import sys
sys.setrecursionlimit(10**6)

from bisect import bisect_left
from collections import deque
"""
行き ⇨ dp LIS計算とstack登録
帰り ⇨ stackをpop dp巻き戻し
"""

n = int(input())
As = list(map(int, input().split()))
edges = [[] for i in range(n)]
for i in range(n-1):
    u,v = map(int, input().split())
    u -= 1
    v -= 1
    edges[u].append(v)
    edges[v].append(u)


INF = 10**10
dp = []
pos = 0
ans = [0] * n
q = deque([])
def dfs(v,p):
    global pos
    # 行き
    a = As[v]
    ind = bisect_left(dp, a)
    if ind == len(dp): 
        dp.append(a)
        # 引数の1つ目はappendしたか
        q.append((True, -1, -1))
    else:
        # 更新
        tmp = dp[ind]
        dp[ind] = a
        q.append((False,ind,tmp))

    ans[v] = len(dp)
    for u in edges[v]:
        if u == p:continue
        dfs(u,v)
        # 帰り        
        isappend, ind, tmp = q.pop()
        if isappend:
            dp.pop()
        else:
            dp[ind] = tmp
        
    
dfs(0,-1)

print(*ans)