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 <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 
#define FIO ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define PI 3.141592653589793238462643383279502884L
#define make_unique(vec) vec.resize(distance(vec.begin(),unique(vec.begin(),vec.end())))
#define Sort(vec) sort(vec.begin(),vec.end())
#define RSort(vec) sort(vec.rbegin(),vec.rend())
#define endl "\n"
#define tr1(i) cout<<i<<endl;
#define tr2(i,j) cout<<i<<" "<<j<<endl;
#define tr3(i,j,k) cout<<i<<" "<<j<<" "<<k<<endl;
#define vvi vector<vector<int> > 
#define vvl vector<vector<ll> >
#define all(st) st.begin(),st.end()
#define vpl vector<pair<ll,ll> >
#define vpi vector<pair<int,int> >
#define pi pair<int,int>
#define pl pair<ll,ll>
#define al vector<list<int> >
#define vs vector<string>
#define vb vector<bool>
#define vi vector<int>
#define vl vector<ll>
#define vc vector<char>
#define rep(i,l,r) for(int i=l;i<r;i++)
#define repit(st) for(auto it=st.begin();it!=st.end();it++)
#define cs(ans); cout<<setprecision(6)<<fixed<<ans<<endl;
#define ull unsigned long long int
#define eb emplace_back
#define pb push_back
#define ll long long int
#define minf -(1e18)
#define inf 1e18
#define f first
#define se second
#define mod 1000000007
using namespace std;
using namespace __gnu_pbds; 
ll mi(ll n,ll m){ll pw=n%m;ll ex=m-2;ll ans=1;while(ex){if(ex&1) ans = ans*pw%m;pw = pw*pw%m;ex>>=1;}return ans%mod;}
ll pw(ll a,ll n){ll pw=a;ll ex=n;ll ans=1;while(ex){if(ex&1) ans = ans*pw;pw = pw*pw;ex>>=1;}return ans;}
ll pwm(ll a,ll n){ll pw=a%mod;ll ex=n;ll ans=1;while(ex){if(ex&1) ans = ans*pw%mod;pw = pw*pw%mod;ex>>=1;}return ans%mod;}
template<typename T>istream &operator>>(istream &is, vector<T> &v) { for (T &t : v) is >> t; return is; }
template<typename T>ostream &operator<<(ostream &os, const vector<T> &v) { for (const T &t : v) {os << t <<" ";} os << '\n'; return os; }
template<typename T> using ordered_set=tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
//st.order_of_key(num)->number of elements < num
//st.find_by_order(i)->iterator of ith element in sorted order starting from zero
ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll max(ll a,ll b){return a>b?a:b;}
ll min(ll a,ll b){return a<b?a:b;}
ll Abs(ll x){return x>0?x:(-x);}
ll Ceil(ll a,ll b){return a/b+(a%b!=0);}
#define bs binary_search
#define lb lower_bound
#define ub upper_bound
#define mkp make_pair
//#define int long long int
//Binary_Search->dp->greedy
//auto check = [&](ll x){};
void go(int i,int par,vl&ans,set<int> st,vl&vec,vvi&adj){
	if(st.lb(vec[i])==st.end()) st.insert(vec[i]),ans[i]=st.size();
	else{
		ans[i]=st.size();
		int val=*st.lb(vec[i]);
		st.erase(val);
		st.insert(vec[i]);
	}
	for(int nbs:adj[i]){
		if(nbs!=par){
			go(nbs,i,ans,st,vec,adj);
		}
	}
}
	
void solve(){
	int n;
	cin>>n;
	vl vec(n);
	cin>>vec;
	vvi adj(n);
	for(int i=0;i<n-1;i++){
		int x,y;
		cin>>x>>y;
		x--;y--;
		adj[x].pb(y);
		adj[y].pb(x);
	}
	vl ans(n);
	set<int> st;
	go(0,-1,ans,st,vec,adj);
	for(int i=0;i<n;i++) cout<<ans[i]<<endl;
		
	
	
	
		
	
}
int32_t main(){
int t=1;
//scanf("%d",&t);
while(t--){
	solve();
}
return 0;
}