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 bisect as bs
import sys

sys.setrecursionlimit(2*10**5+1)

def dfs(i, nbr, A, prv, done, history, B, ans):#深さ優先探索（再帰）
  
  j = bs.bisect_left(B,A[i]) #Bへの挿入予定位置
  #print(i,A[i],j,B)
  
  #A[i]をhistoryの適当な位置に挿入
  if j < len(B):
    history[i] = (B[j], A[i], j) # 更新前の値、更新後の値、更新位置
    B[j] = A[i]
  else:
    history[i] = (-1, A[i], j) # 「追加する」 、追加する値、追加位置（最後）
    B.append(A[i])
  
  ans[i] = len(B) #答え書き込み
  done[i] = 1 #探索済み
  #print(i,B)
  
  for a in nbr[i]:
    if done[a] == 0: #探索前のみ
      dfs(a, nbr, A, prv, done, history, B, ans)
      prv[a] = i
      
  j = history[i][2]
  if history[i][0] >= 0:
    B[j] = history[i][0] #Bを現状復帰
  else:
    B.pop()
    
N = int(input()) 
A = list(map(int,input().split()))

nbr = [[] for _ in range(N)]

for _ in range(N-1):
  u,v = map(int,input().split())
  u -= 1
  v -= 1
  nbr[u].append(v)
  nbr[v].append(u)

#print(nbr)
  
prv = [-1 for _ in range(N)]
done = [0 for _ in range(N)] 
history = [0 for _ in range(N)]
B = []
ans = [0 for _ in range(N)]

dfs(0, nbr, A, prv, done, history, B, ans)

for i in range(N):
  print(ans[i])
