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

N=int(input())
*A,=map(int,input().split())
INF=10**20
G=[[] for _ in range(N)]
ab = [tuple(map(int,input().split())) for _ in range(N-1)]
for a,b in ab:
  G[a-1].append(b-1)
  G[b-1].append(a-1)
  
def dfs(a0):
    seen =[0]*len(G)
    todo = [~a0, a0]
    while todo:
        a = todo.pop()
        if a >= 0:
            seen[a] = 1
            idx = bisect_left(lis,A[a])
            his[a] = (idx,lis[idx])
            lis[idx] = A[a]
            dp[a] = bisect_left(lis,INF)
            for b in G[a]:
                if seen[b]: continue
                todo.append(~b)
                todo.append(b)
        else:
          idx,val = his[~a]
          lis[idx] = val
    return
  
lis = [INF]*N
dp = [0]*N
his = {}
dfs(0)
print(*dp,sep="\n")