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 C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
const int asize=2e5+5;
vector<int> adjl[asize],lv[asize];
int val[asize],lis[asize],n;

void dfs(int pos,int par){
	for(auto x:adjl[pos]){
		if(x!=par){
			lv[x]=lv[pos];
			int c=val[x];
			auto it=lower_bound(lv[x].begin(),lv[x].end(),c);
			if(it==lv[x].end())lv[x].push_back(c);
			else *it=c;
			lis[x]=lv[x].size();
			if(adjl[x].size()>1)dfs(x,pos);
		}
	}
}

int main(){
	ios_base::sync_with_stdio(0),cin.tie(NULL);
	int a,b;
	vector<int>aaa;
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>val[i];
	}
	for(int i=0;i<n-1;i++){
		cin>>a>>b;
		adjl[a].push_back(b);
		adjl[b].push_back(a);
	}
	aaa.push_back(val[1]);
	lv[1]=aaa;
	lis[1]=1;
	dfs(1,-1);
	for(int i=1;i<=n;i++){
		cout<<lis[i]<<"\n";
	}
	return 0;
}