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: Wrong Answer)

#include<bits/stdc++.h>
#define ll          long long
#define pb          push_back
#define	endl		'\n'
#define pii         pair<ll int,ll int>
#define vi          vector<ll int>
#define all(a)      (a).begin(),(a).end()
#define F           first
#define S           second
#define sz(x)       (ll int)x.size()
#define hell        1000000007
#define rep(i,a,b)	for(ll int i=a;i<b;i++)
#define lbnd        lower_bound
#define ubnd        upper_bound
#define bs          binary_search
#define mp          make_pair
using namespace std;

#define N  200005

vi adj[N];
ll a[N];
ll ans[N];
vi u;
map<ll,ll> mx;

void dfs(int node,int par){
	int fl=-1,last=-1;
	if(mx[a[node]]>0){
		ans[node]=ans[mx[a[node]]];
	}
	else{
		mx[a[node]]=node;
		if(sz(u)==0){
			u.pb(a[node]);
			ans[node]=1;
			fl=1;
		}
		else{
			int pos=ubnd(all(u),a[node])-u.begin();
			if(pos==sz(u)){
				u.pb(a[node]);
				ans[node]=pos+1;
				fl=1;
			}
			else{
				last=u[pos];
				u[pos]=a[node];
				ans[node]=pos+1;
				fl=2;
			}
		}
	}
	for(auto it:adj[node]){
		if(it!=par){
			dfs(it,node);
		}
	}
	if(fl==1){
		u.pop_back();
		mx[a[node]]=0;
	}
	if(fl==2){
		mx[a[node]]=0;
		int pos=lbnd(all(u),a[node])-u.begin();
		u[pos]=last;
	}
}

void solve()
{
	int n;
	cin>>n;
	rep(i,1,n+1)
		cin>>a[i];
	rep(i,1,n){
		int a,b;
		cin>>a>>b;
		adj[a].pb(b);
		adj[b].pb(a);
	}
	dfs(1,0);
	rep(i,1,n+1)
		cout<<ans[i]<<endl;
}

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int TESTS=1;
//	cin>>TESTS;
	while(TESTS--)
	{
		solve();
	}
	return 0;
}