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
from bisect import bisect_left
sys.setrecursionlimit(10**6)
N,*L = map(int, open(0).read().split())
A = [0]+L[:N]
dic = [[] for i in range(N+1)]
INF = 10**20
dp = [INF]*N
ans = [0]*(N+1)
for u,v in zip(*[iter(L[N:])]*2):
  dic[u].append(v)
  dic[v].append(u)
def dfs(v,p):
  i = bisect_left(dp,A[v])
  old = dp[i]
  dp[i] = A[v]
  ans[v] = bisect_left(dp,INF)
  for u in dic[v]:
    if u==p:
      continue
    dfs(u,v)
  dp[i] = old
dfs(1,-1)
for i in range(1,N+1):
  print(ans[i])