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;
#define MOD 1000000007
//#define MOD 998244353
#define INF 1000000010
#define EPS 1e-9
#define F first
#define S second

#define debug(x) cout<<x<<endl;
#define repi(i,x,n) for(int i=x;i<n;i++)
#define rep(i,n) repi(i,0,n)
#define lp(i,n) repi(i,0,n)
#define repn(i,n) for(int i=n;i>=0;i--)
#define int long long
#define endl "\n"

typedef pair<int,int> PII;
typedef pair<int,string> PIS;
typedef pair<string,int> PSI;


signed main(){
  cin.tie(0);	
  // ios::sync_with_stdio(false);
  int n;
  cin>>n;
  int a[n];
  rep(i,n) cin>>a[i];
  vector<int> nod[n];
  rep(i,n-1){
    int a,b;
    cin>>a>>b;
    a--;b--;
    nod[a].push_back(b);
    nod[b].push_back(a);
  }
  queue<pair<int,vector<int> > > bfs;
  vector<int> ans[n];
  int flag[n]={0};
  vector<int> ini;
  ini.push_back(a[0]);
  flag[0]=1;
  bfs.push(make_pair(0,ini));
  while(!bfs.empty() ){
    pair<int,vector<int> > p;
    p=bfs.front();
    bfs.pop();
    ans[p.F]=p.S;
    rep(i,nod[p.F].size() ){
      if(flag[nod[p.F][i]]==0){
	pair<int,vector<int> > q=p;
	flag[nod[p.F][i]]=1;
	q.S.push_back(a[nod[p.F][i]]);
	bfs.push(make_pair(nod[p.F][i],q.S));
      }
    }
  }

  vector<int> aa;
  rep(i,n){
    aa.clear();
    //cout<<i<<endl;
    rep(j,ans[i].size() ){
      auto itr=lower_bound(aa.begin(),aa.end(),ans[i][j]);
      if(itr==aa.end() ) aa.push_back(ans[i][j]);
      else aa[itr-aa.begin()]=ans[i][j];
      //cout<<j<<endl;
    }
    cout<<aa.size()<<endl;
  }
  return 0;
}