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
import sys
sys.setrecursionlimit(500000)

def LIS(lis):

    seq = []

    for c in lis:
        ind = bisect.bisect_left(seq,c)

        if ind == len(seq):
            seq.append(c)
        else:
            seq[ind] = c

    return len(seq)


N = int(input())

a = list(map(int,input().split()))

lis = [ [] for i in range(N) ]

for i in range(N-1):

    u,v = map(int,input().split())
    u -= 1
    v -= 1

    lis[u].append(v)
    lis[v].append(u)

ans = [None] * N

def dfs(v,p,ln):

    ind = bisect.bisect_left(ln,a[v])

    if ind == len(ln):
        mt = None
        ln.append(a[v])
    else:
        mt = ln[ind]
        ln[ind] = a[v]

    ans[v] = len(ln)

    for nex in lis[v]:

        if nex != p:
            dfs(nex,v,ln)

    if mt == None:
        del ln[-1]
    else:
        ln[ind] = mt

    return

dfs(0,0,[])

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