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>
#include <iostream>
using namespace std;
typedef int64_t llo;
#define mp make_pair
#define a first
#define b second
#define pb push_back
int ma=0;
int ss[200001];
int ans[200001];
int it[200001];
vector<int> adj[200001];
int dfs(int no,int par=-1){
	pair<int,pair<int,int>> rev;
	if(ma==0){
		ss[0]=it[no];
		ma+=1;
		rev={-1,{-1,-1}};
	}
	else if(it[no]>ss[ma-1]){
		ma+=1;
		ss[ma-1]=it[no];
		rev={-1,{-1,-1}};
	}
	else if(it[no]==ss[ma-1]){
		
	}
	else{
		int low=0;
		int high=ma-1;
		while(low>high-1){
			int mid=(low+high)/2;
			if(ss[mid]>it[no]){
				high=mid;
			}
			else{
				low=mid+1;
			}
		}
		int ind=high;
		if(ss[low]>it[no]){
			ind=min(ind,low);
		}
		rev={ind,{ss[ind],it[no]}};
		ss[ind]=it[no];
	}
	ans[no]=ma;
	for(auto j:adj[no]){
		if(j==par){
			continue;
		}
		dfs(j,no);
	}
	if(rev.a!=-1){
		ss[rev.a]=rev.b.a;
	}
	else{
		ma-=1;
	}
}
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	int n;
	cin>>n;
	
	for(int i=0;i<n;i++){
		cin>>it[i];
	}
	int aa,bb;
	for(int i=0;i<n-1;i++){
		cin>>aa>>bb;
		adj[aa-1].pb(bb-1);
		adj[bb-1].pb(aa-1);
	}
	dfs(0,-1);
	for(int i=0;i<n;i++){
		cout<<ans[i]<<endl;
	}

	return 0;
}