Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

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

Sample Output 1

1
2
3
3
4
4
5
2
2
3

For example, the sequence A obtained from the shortest path from Vertex 1 to Vertex 5 is 1,2,5,3,4. Its longest increasing subsequence is A_1, A_2, A_4, A_5, with the length of 4.

Solution in Julia (1.4.0) (judge verdict: Wrong Answer)

parseInt(x)=parse(Int, x)
parseMap(x::Array{SubString{String},1})=map(parseInt, x)
function bsearch(a::Array{Int,1},x);l,r=0,length(a)+1;while r-l>1;m=(r+l)>>1;if a[m]<x;l=m;else;r=m;end;end;l;end

function main()
	n = readline() |> parseInt
	a = readline() |> split |> parseMap
	e = [Int[] for i in 1:n]
	for i in 1:n-1
		u,v = readline() |> split |> parseMap
		push!(e[u],v)
		push!(e[v],u)
	end
	t = zeros(Int,n)
	ans = zeros(Int,n)
	lis = 10^18*ones(Int,n)
	bef=Tuple{Int,Int}[(0,0) for i in 1:n]
	sf = zeros(Int,n)
	sf[1]=1
	t[1]=1
	ans[1]=1
	lis[1]=a[1]
	bef[1]=(1,a[1])
	stack = Int[1]
	while !isempty(stack)
		now = pop!(stack)
		if sf[now]==0
			push!(stack,t[now])
			sf[now]=1
			k = bsearch(lis,a[now])+1
			bef[now]=(k,lis[k])
			lis[k]=a[now]
			ans[now]=k
		end
		f = 0
		for i in e[now]
			if t[i]==0
				f = 1
				t[i]=now
				push!(stack,i)
				break
			end
		end
		if f == 0
			k = bef[now][1]
			lis[k]=bef[now][2]
		end
	end
	for i in 1:n
		println(ans[i])
	end
end

main()